Circles – JEE Main MathematicsPractice Questions & PYQs

The circle passing through the origin $O$, point $A$ on the x-axis, and point $B$ on the y-axis forms a right-angled triangle with the diameter being the hypotenuse $AB$. Given the radius is 4, the length of this diameter is 8, which leads to the relationship $AB^2 = 64$. Applying the distance formula to coordinates $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, this becomes $3a^2 + 2b^2 = 64$.

The centroid $(x, y)$ of triangle $OAB$ has coordinates $x = -\frac{\sqrt{3}a}{3}$ and $y = -\frac{\sqrt{2}b}{3}$. Rearranging these to solve for $a$ and $b$ yields $a = -\sqrt{3}x$ and $b = -\frac{3y}{\sqrt{2}}$. Substituting these values into the previous equation gives $3(-\sqrt{3}x)^2 + 2\left(-\frac{3y}{\sqrt{2}}\right)^2 = 64$, which simplifies to $9x^2 + 9y^2 = 64$. Dividing by 9 results in $x^2 + y^2 = \left(\frac{8}{3}\right)^2$, identifying the locus of the centroid as a circle with radius $\frac{8}{3}$.

When chords are drawn from a point $P(1, 2)$ to a circle given by $x^{2}+y^{2}+x-3y=0$ and are bisected by the y-axis, the midpoint of each chord must lie on the line $x=0$. For any chord with an endpoint $(x, y)$, the midpoint of the segment connecting $P(1, 2)$ to $(x, y)$ is given by $\left(\frac{x+1}{2}, \frac{y+2}{2}\right)$. Since this midpoint lies on the y-axis, its x-coordinate must be zero, implying $\frac{x+1}{2}=0$, which determines that the x-coordinate of the endpoints is $x=-1$.

Substituting $x=-1$ into the circle's equation $x^{2}+y^{2}+x-3y=0$ allows us to find the corresponding y-coordinates of the chord endpoints. Plugging in $-1$ results in $(-1)^{2}+y^{2}-1-3y=0$, which simplifies to $y^{2}-3y=0$. Solving this quadratic equation gives $y(y-3)=0$, indicating that the y-coordinates are $0$ and $3$. Consequently, the endpoints of the chords are $R(-1, 0)$ and $S(-1, 3)$.

The midpoint $(\alpha, \beta)$ of the line segment $RS$ is found by averaging the coordinates of $R$ and $S$, which yields $\left(\frac{-1-1}{2}, \frac{0+3}{2}\right)$ or $(-1, 1.5)$. Evaluating the required expression $6(\alpha+\beta)$ with $\alpha=-1$ and $\beta=1.5$ results in $6(-1+1.5)$, which equals $3$.

To find the locus of the foot of the perpendicular $(h, k)$ drawn from the origin to a chord $AB$, we first note that the equation of this chord is given by $hx + ky = h^2 + k^2$, which can be rewritten as $\frac{hx+ky}{h^2+k^2} = 1$. This chord intersects the circle $x^2 + y^2 - 6x - 8y - 11 = 0$. By homogenizing the circle equation with the chord equation, we obtain the pair of lines $OA$ and $OB$ connecting the origin to the endpoints of the chord. This process involves replacing the constant term with $1^2$ and the linear terms with the linear expression multiplied by $1$, resulting in $x^2 + y^2 - (6x + 8y) \left( \frac{hx+ky}{h^2+k^2} \right) - 11 \left( \frac{hx+ky}{h^2+k^2} \right)^2 = 0$.

Since the chord subtends a right angle at the origin, the lines $OA$ and $OB$ are perpendicular, which requires the sum of the coefficients of $x^2$ and $y^2$ in the homogenized equation to be zero. Extracting these coefficients from the expression, we find the contribution from the terms $x^2$ and $y^2$ is $1 + 1 = 2$, from the cross-terms is $-\frac{6h+8k}{h^2+k^2}$, and from the squared term is $-\frac{11}{h^2+k^2}$. Setting the sum $2 - \frac{6h+8k}{h^2+k^2} - \frac{11}{h^2+k^2} = 0$ gives the relation $2(h^2 + k^2) - 6h - 8k - 11 = 0$, or $h^2 + k^2 - 3h - 4k - 5.5 = 0$. Replacing $h$ and $k$ with $x$ and $y$ provides the locus $x^2 + y^2 - 3x - 4y - 5.5 = 0$. Comparing this with $x^2 + y^2 - \alpha x - \beta y - \gamma = 0$ yields $\alpha = 3$, $\beta = 4$, and $\gamma = 5.5$, making the value of $\alpha + \beta + 2\gamma$ equal to $3 + 4 + 2(5.5)$, which results in 18.

Any chord of a circle that passes through a specific interior point is farthest from the center if it is perpendicular to the radius connecting the center to that point. Circle $C_1$ has a diameter of 10 and lies in the region $x \ge 0$ while passing through the origin, which places its center at $(5, 0)$ and establishes the equation $(x-5)^2 + y^2 = 25$. The chord $y = x$ intersects this circle at the points $(0, 0)$ and $(5, 5)$, which serve as the endpoints for the diameter of circle $C_2$, locating the center $M$ of $C_2$ at $(2.5, 2.5)$.

Identifying the required chord relies on calculating the slope of the line segment $MP$ connecting $(2.5, 2.5)$ to the point $(2, 3)$, which gives $m = -1$. Because the chord must be perpendicular to $MP$, its slope is 1, leading to the line equation $x - y + 1 = 0$. Comparing this result to the given expression $x + ay + b = 0$, the coefficients are $a = -1$ and $b = 1$, which means $a - b = -2$.

A circle defined by its center at the intersection of two perpendicular lines is effectively determined by the relationship between its radius and the perpendicular distance to any intersecting chord. Since the lines are perpendicular, the sum of the products of their coefficients $A_1A_2 + B_1B_2$ must equal zero, establishing the geometric constraints necessary to locate the center. Once the center and radius are known through the circle's passage through the origin, the length of any chord is derived from the Pythagorean relationship between the radius, the chord's distance from the center, and half the chord length.

The condition $1(2) + (k - 1)k^2 = 0$ must be met for the lines to be perpendicular, resulting in the cubic equation $k^3 - k^2 + 2 = 0$. Testing integer values shows that $k = -1$ satisfies this relationship. Substituting this value into the original line equations yields $x - 2y + 3 = 0$ and $2x + y - 4 = 0$. Solving this system of equations simultaneously identifies the center of the circle at the point $(1, 2)$.

Since the circle includes the origin, the squared radius $r^2$ equals the squared distance from $(1, 2)$ to $(0, 0)$, which is $1^2 + 2^2 = 5$. The perpendicular distance $d$ from the center $(1, 2)$ to the line $x - y + 2 = 0$ is calculated as $d = \frac{|1 - 2 + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$. Utilizing the geometric relationship $r^2 = d^2 + \left(\frac{AB}{2}\right)^2$, the equation becomes $5 = \frac{1}{2} + \frac{(AB)^2}{4}$. Solving for the squared chord length yields $\frac{9}{2} = \frac{(AB)^2}{4}$, which gives a final value of $18$.

The concept of a perpendicular bisector relies on the property that any point on such a line is equidistant from the two endpoints of the segment it bisects. Consequently, the midpoint of the segment $PQ$ must lie on the given perpendicular bisector $x-y+1=0$, and the slope of the segment $PQ$ must be the negative reciprocal of the slope of that bisector.

Defining $P$ as $(x_1, y_1)$ and $Q$ as $(x_2, y_2)$, the bisector $x-y+1=0$ has a slope of $1$, which forces the slope of $PQ$ to be $-1$. This relationship, $\frac{y_2 - y_1}{x_2 - x_1} = -1$, simplifies to $y_2 = x_1 + y_1 - x_2$. Given that the midpoint $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ lies on $x-y+1=0$, we have $\frac{x_1+x_2}{2} - \frac{y_1+y_2}{2} + 1 = 0$, which can be rearranged to $x_1 + x_2 - y_1 - y_2 + 2 = 0$. Since $Q$ also satisfies $5x_2 + y_2 + 2 = 0$, we can express $y_2$ as $-5x_2 - 2$. Substituting this into the midpoint equation yields $x_1 + x_2 - y_1 - (-5x_2 - 2) + 2 = 0$, or $x_1 + 6x_2 - y_1 + 4 = 0$. From the slope condition and the line equation for $Q$, $x_2$ is determined as $\frac{-x_1 - y_1 - 2}{4}$. Substituting this value of $x_2$ back into the simplified midpoint equation results in $x_1 + 6\left(\frac{-x_1 - y_1 - 2}{4}\right) - y_1 + 4 = 0$, which simplifies to $x_1 = 2 - 5y_1$.

Because $P$ lies on the circle $x^2+y^2=4$, substituting $x_1 = 2 - 5y_1$ into the circle equation gives $(2 - 5y_1)^2 + y_1^2 = 4$. Expanding this provides $4 - 20y_1 + 25y_1^2 + y_1^2 = 4$, which leads to $26y_1^2 - 20y_1 = 0$. Factoring this quadratic equation, $2y_1(13y_1 - 10) = 0$, reveals the possible $y_1$ values of $0$ and $\frac{10}{13}$. Using the relation $x_1 = 2 - 5y_1$, the corresponding abscissae for $P$ are $2$ and $-\frac{24}{13}$. Adding these two values yields a sum of $\frac{2}{13}$. Finally, multiplying this sum by $13$ results in $2$.

The geometric condition $PQ=PR$ implies that point $P$ must lie on the perpendicular bisector of the chord formed by the intersection of the line and the circle. Since the perpendicular bisector of any chord in a circle necessarily passes through the circle's center $(4, -3)$, this bisector acts as a diameter of the circle. Given the chord $x-y=4$ has a slope of $1$, the perpendicular diameter must possess a slope of $-1$. Passing through the center $(4, -3)$, the equation for this diameter becomes $y - (-3) = -1(x - 4)$, which simplifies to $x + y = 1$.

We can find the coordinates of $P(\alpha, \beta)$ by identifying the intersection of this diameter with the circle $C$. Using the parametric form for a circle with radius $3$ and an angle of $135^{\circ}$ relative to the positive x-axis for a line with slope $-1$, the coordinates are $\alpha = 4 + 3\cos(135^{\circ})$ and $\beta = -3 + 3\sin(135^{\circ})$, which simplifies to $\alpha = 4 - \frac{3}{\sqrt{2}}$ and $\beta = -3 + \frac{3}{\sqrt{2}}$. Substituting these values into the required expression $(6\alpha + 8\beta)^2$ yields $(6(4 - \frac{3}{\sqrt{2}}) + 8(-3 + \frac{3}{\sqrt{2}}))^2$, which further simplifies to $(24 - \frac{18}{\sqrt{2}} - 24 + \frac{24}{\sqrt{2}})^2$. Squaring $(\frac{6}{\sqrt{2}})^2$ results in $18$.

The maximum distance from an external point to a circle is determined by extending the line segment connecting that point to the center of the circle until it intersects the far side of the circumference, making the total distance equal to the distance between the point and the center plus the radius. We first determine the circle's properties by rewriting the equation $x^{2}+y^{2}-6x-8y+21=0$ in the standard form $(x-3)^{2}+(y-4)^{2}=4$, which places the center at $(3, 4)$ and gives a radius $r=2$.

For the parabola $x^{2}+6x+y+13=0$, we complete the square for the $x$ terms to get $(x+3)^{2} = -y-4$, which simplifies to $(x+3)^{2} = -(y+4)$. This identifies the vertex of the parabola at $(-3, -4)$. Calculating the distance between the vertex $(-3, -4)$ and the circle's center $(3, 4)$ yields $\sqrt{(3 - (-3))^{2} + (4 - (-4))^{2}} = \sqrt{6^{2} + 8^{2}} = 10$. Adding the radius of the circle to this value results in $10 + 2 = 12$ as the required maximum distance.

The vertex $P$ of the parabola $y = x^2 - 6x + 12$ is found by completing the square, giving $y = (x-3)^2 + 3$, which identifies $P$ as $(3, 3)$. The circle $x^2 + y^2 - 2x - 4y + 3 = 0$ can be rewritten as $(x-1)^2 + (y-2)^2 = 2$, establishing a center $C$ at $(1, 2)$ and a radius $r = \sqrt{2}$. For a line passing through $P$ and intersecting the circle at $R$ and $S$, the distance sum $PR + PS$ is maximized when the line passes through the circle's center $C$, effectively making the chord $RS$ a diameter. In this orientation, the distances are $PC - r$ and $PC + r$, which simplify their sum to $2PC$. Given $PC^2 = (3-1)^2 + (3-2)^2 = 5$, the maximum value is $4 \times 5 = 20$.

The geometry of tangents from an external point dictates that the angle between the two tangents is supplementary to the central angle subtended by the points of contact. For the given circle $x^2+y^2-4x-6y-3=0$, the centre $O$ is at $(2,3)$ and the radius $r$ is calculated as $\sqrt{2^2+3^2-(-3)} = \sqrt{16} = 4$. Given that the central angle $\angle AOB$ is $\frac{\pi}{3}$, or $60^\circ$, the angle between the tangents at their intersection point $P(x,y)$ must be $180^\circ - 60^\circ = 120^\circ$.

Bisecting this angle creates a right-angled triangle at the point of tangency, where the tangent of the half-angle is the ratio of the radius to the length of the tangent segment. With the half-angle equal to $60^\circ$, the relationship is $\tan 60^\circ = \frac{r}{\sqrt{d^2-r^2}}$, where $d$ is the distance from the centre $(2,3)$ to the external point $P(x,y)$. Squaring the expression $\sqrt{3} = \frac{4}{\sqrt{d^2-16}}$ results in $3 = \frac{16}{d^2-16}$, which simplifies to $3(d^2-16) = 16$. Substituting $d^2 = (x-2)^2 + (y-3)^2$ back into this equation yields $3((x-2)^2 + (y-3)^2 - 16) = 16$. Expanding this expression leads to $3(x^2+y^2-4x-6y-3) = 16$, which simplifies to the final locus equation $3(x^2+y^2)-12x-18y-25=0$.

The circle $x^2+y^2=4$ is centered at the origin with a radius of $2$ and intersects the $x$-axis at $A(2, 0)$ and $B(-2, 0)$. The line $BP$ passing through $B(-2, 0)$ and $P(2\cos\alpha, 2\sin\alpha)$ has a slope of $\frac{2\sin\alpha}{2\cos\alpha+2} = \tan(\alpha/2)$, resulting in the equation $y = \tan(\alpha/2)(x+2)$. Similarly, the line $AQ$ passing through $A(2, 0)$ and $Q(2\cos\beta, 2\sin\beta)$ has a slope of $\frac{2\sin\beta}{2\cos\beta-2} = -\cot(\beta/2)$, yielding the equation $y = -\cot(\beta/2)(x-2)$.

By denoting the intersection point of these two lines as $(x, y)$, we obtain the relations $\tan(\alpha/2) = \frac{y}{x+2}$ and $\tan(\beta/2) = \frac{2-x}{y}$. Given the condition $\alpha - \beta = \pi/2$, it follows that $\frac{\alpha}{2} - \frac{\beta}{2} = \pi/4$. Applying the tangent subtraction identity to this angular relation gives $\frac{\tan(\alpha/2) - \tan(\beta/2)}{1 + \tan(\alpha/2)\tan(\beta/2)} = 1$. Substituting the expressions for $\tan(\alpha/2)$ and $\tan(\beta/2)$ into this identity yields $\frac{\frac{y}{x+2} - \frac{2-x}{y}}{1 + (\frac{y}{x+2})(\frac{2-x}{y})} = 1$, which simplifies to $\frac{y^2+x^2-4}{4y} = 1$. This equation rearranges to the final form $x^2+y^2-4y-4=0$.

For two circles to intersect at exactly two distinct points, the distance between their centers must be strictly greater than the absolute difference of their radii and strictly less than their sum. The equation of the first circle, $(x+1)^2+(y+4)^2=r^2$, identifies a center at $C_1=(-1,-4)$ and a radius of $r$, whereas the second equation, $x^2+y^2-4x-2y-4=0$, completes to $(x-2)^2+(y-1)^2=9$, indicating a center at $C_2=(2,1)$ with a radius of $3$. The distance $d$ between these two centers is determined by $d=\sqrt{(2-(-1))^2+(1-(-4))^2}=\sqrt{3^2+5^2}=\sqrt{34}$.

By the geometric condition for intersection, $|r-3| < \sqrt{34} < r+3$. From the inequality $\sqrt{34} < r+3$, it follows that $r > \sqrt{34}-3$. The inequality $|r-3| < \sqrt{34}$ further constrains $r$ within the range $3-\sqrt{34} < r < 3+\sqrt{34}$. Because a radius must be positive and $3-\sqrt{34}$ is negative, the valid interval for $r$ is $(\sqrt{34}-3, \sqrt{34}+3)$. Identifying these boundaries as $\alpha$ and $\beta$ allows for the product $\alpha \beta = (\sqrt{34}-3)(\sqrt{34}+3)$, which resolves to $34-9$, yielding 25.

The line perpendicular to chord $AB$ that passes through its midpoint must inherently pass through the center of the circle, which is the origin $(0,0)$ for the circle $x^2+y^2=4$. Since the line intersects the circle at $C$ and $D$, the segment $CD$ functions as a diameter, giving it a length of $2 \times 2 = 4$. To find the length of the chord $AB$, we calculate the perpendicular distance $d$ from the origin to the line $x+y=1$ using the formula $d = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$, which yields $d = \frac{|-1|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$. Applying the chord length formula $2\sqrt{r^2-d^2}$, we obtain $AB = 2\sqrt{4 - \frac{1}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{14}$. Because the diagonals $AB$ and $CD$ are perpendicular, the area of the quadrilateral $ADBC$ is equal to half the product of its diagonals, resulting in $\frac{1}{2} \times 4 \times \sqrt{14} = 2\sqrt{14}$.

Since the lines $x+y=3$ and $x-y=3$ are perpendicular and intersect at the point $(3,0)$, any circle touching both must have its center located on one of the angle bisectors. By verifying the distance requirements for the point $(-9,4)$, we find the centers must lie on the x-axis, allowing the center to be represented as $(\alpha, 0)$.

The radius $r$ of the circle satisfies the condition $r = | \alpha-3 | / \sqrt{2}$ based on the distance to the lines, while the distance to the point $(-9,4)$ requires $r^2 = (\alpha+9)^2 + 16$. Equating these expressions gives $(\alpha-3)^2 / 2 = (\alpha+9)^2 + 16$, which simplifies to the quadratic equation $\alpha^2 + 42\alpha + 185 = 0$.

Solving for the center coordinate yields $\alpha = -5$ and $\alpha = -37$. Calculating the squared radii for these positions using $r^2 = (\alpha-3)^2 / 2$ results in 32 and 800. The absolute difference between these two squared radii is 768.

A circle determined by the diameter endpoints $(0,0)$ and $(4,6)$ follows the equation $x(x-4) + y(y-6) = 0$, which simplifies to $x^2 - 4x + y^2 - 6y = 0$. By rearranging this into standard form, we identify the center at $(2, 3)$ and a radius squared of $r^2 = 13$. For the point $(k, 3k)$ to lie on this circle, its coordinates must satisfy the equation $k^2 - 4k + (3k)^2 - 6(3k) = 0$, which simplifies to $10k^2 = 22k$. Given that the points are distinct, $k$ must be non-zero, allowing us to solve $10k = 22$ and calculate the final expression $10k + r^2$ as $22 + 13$, resulting in 35.

Two circles intersect at exactly two points if and only if the distance between their centers lies strictly between the absolute difference of their radii and the sum of their radii. Since circle $C$ has a radius of 2 and touches both axes in the second quadrant, its center is located at $(-2, 2)$. The distance $d$ between this center and the given center at $(2, 5)$ is calculated as follows:

$d = \sqrt{(2 - (-2))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = 5$

The condition for having exactly two intersection points is $|r - 2| < d < r + 2$. Substituting the value of $d$, we require $|r - 2| < 5 < r + 2$. The inequality $5 < r + 2$ yields $r > 3$, while the inequality $|r - 2| < 5$ simplifies to $-5 < r - 2 < 5$, which results in $r < 7$. Combining these constraints gives the interval $(3, 7)$ for $r$, where $\alpha = 3$ and $\beta = 7$. Applying these values to the final expression, $3\beta - 2\alpha$ becomes $3(7) - 2(3)$, which equals 15.

When two circles touch externally, the distance between their centres equals the sum of their radii, and the point of contact divides the line segment joining the centres internally in the ratio of their radii. The centre of $C_1$ is $(-3, -3)$, so the distance to the centre $(1, 3)$ of $C_2$ is $\sqrt{16+36}=\sqrt{52}$, which implies the radius of $C_2$ is $r_2 = \sqrt{52}-3$.

Using the section formula to divide the line segment between the centres in the ratio $3 : (\sqrt{52}-3)$, the point of contact $(\alpha, \beta)$ has coordinates $\alpha = \frac{6}{\sqrt{13}}-3$ and $\beta = \frac{9}{\sqrt{13}}-3$. The difference $\beta-\alpha = \frac{3}{\sqrt{13}}$ yields $(\beta-\alpha)^2 = \frac{9}{13}$, and since $\gcd(9, 13)=1$, the sum $m+n$ equals 22.

The centre of a circle lies at the intersection of the perpendicular bisector of any chord and the given locus equation. Since the circle passes through $(4,2)$ and $(0,2)$, its centre must lie on the perpendicular bisector of this segment, which is the vertical line $x=2$ because the $y$-coordinates of both points are identical. Substituting $x=2$ into the equation $3x+2y+2=0$ gives $6+2y+2=0$, resulting in $y=-4$, which establishes the centre at $(2,-4)$.

The radius squared $r^2$ is the distance from the centre $(2,-4)$ to either given point, such as $(0,2)$, calculated as $(0-2)^2 + (2-(-4))^2 = 4+36 = 40$. A chord with midpoint $M(1,2)$ lies at a distance $d$ from the centre, where $d^2 = (2-1)^2 + (-4-2)^2 = 1^2 + (-6)^2 = 37$. By the property that the radius, the distance to the midpoint, and half the chord form a right-angled triangle, the square of half the chord length is $r^2 - d^2 = 40 - 37 = 3$. Therefore, half the chord length is $\sqrt{3}$, making the total chord length $2\sqrt{3}$.

The reflection of a circle across a line maps its center to a new location while keeping the radius unchanged. For the circle defined by $x^2+y^2-2x+4y-4=0$, completing the square reveals a center of $(1, -2)$ and a radius of $3$. Applying the reflection formula for the point $(1, -2)$ across the line $2x-3y+5=0$ determines the new center of circle $C$ to be $(-3, 4)$. Because $A$ lies on this circle such that $OA$ is parallel to the x-axis and to the right of the center, the coordinates of $A$ are $(0, 4)$.

Point $B(\alpha, \beta)$ lies on circle $C$, and the given arc length $AB$ is one-sixth of the total circumference. Since the total perimeter is $2\pi(3) = 6\pi$, the arc length is $\pi$, which corresponds to a central angle of $\frac{\pi}{3}$ or $60^{\circ}$. With the constraint $\beta < 4$, the point $B$ must be positioned below the center line. We determine the coordinates $(\alpha, \beta)$ relative to the center $(-3, 4)$ by rotating $60^{\circ}$ clockwise from the horizontal radius $OA$, giving $\alpha = -3 + 3 \cos(-60^{\circ})$ and $\beta = 4 + 3 \sin(-60^{\circ})$.

Evaluating these trigonometric values, $\alpha = -3 + 3(\frac{1}{2}) = -\frac{3}{2}$ and $\beta = 4 - 3(\frac{\sqrt{3}}{2})$. To find the final expression, we compute $\beta - \sqrt{3} \alpha$, which becomes $(4 - \frac{3\sqrt{3}}{2}) - \sqrt{3}(-\frac{3}{2})$. This simplifies directly to $4 - \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2}$, resulting in $4$.