Circles – JEE Main MathematicsPractice Questions & PYQs

A family of circles with a fixed radius of $5$ and centers lying on the horizontal line $y=2$ can be expressed by the equation $(x-h)^2 + (y-2)^2 = 25$, where $h$ represents the variable x-coordinate of the center. Because there is a single arbitrary parameter $h$, we can derive the governing differential equation by differentiating the expression once with respect to $x$ to eliminate the constant.

The derivative of $(x-h)^2 + (y-2)^2 = 25$ with respect to $x$ is $2(x-h) + 2(y-2) \frac{dy}{dx} = 0$, which provides the identity $x-h = -(y-2) \frac{dy}{dx}$. Substituting this expression back into the original equation yields $(-(y-2) \frac{dy}{dx})^2 + (y-2)^2 = 25$. Expanding the squared term leads to $(y-2)^2 (\frac{dy}{dx})^2 + (y-2)^2 = 25$, and subtracting $(y-2)^2$ from both sides results in $(y-2)^2 (\frac{dy}{dx})^2 = 25 - (y-2)^2$.

A circle tangent to the $x$-axis with center at $(h, k)$ must have a radius equal to the absolute value of the vertical coordinate of its center, resulting in the equation $(x-h)^2 + (y-k)^2 = k^2$. Since the circle must pass through the point $(-1, 1)$, substituting these coordinates into the equation yields $(-1-h)^2 + (1-k)^2 = k^2$. Expanding this expression results in $h^2 + 2h + 1 + 1 - 2k + k^2 = k^2$, which simplifies to the quadratic equation $h^2 + 2h + 2 - 2k = 0$.

For such a circle to exist, $h$ must be a real value, which requires the discriminant of this quadratic equation to be non-negative. Setting the discriminant $b^2 - 4ac \ge 0$ with $a=1$, $b=2$, and $c=2-2k$ gives $2^2 - 4(1)(2-2k) \ge 0$. Solving this inequality leads to $4 - 8 + 8k \ge 0$, which simplifies to $8k \ge 4$, establishing that $k \ge \frac{1}{2}$.

A chord in a circle of radius 3 subtending an angle of $\frac{2\pi}{3}$ at the center has a midpoint that maintains a constant perpendicular distance from the center. Drawing a line from the center to this midpoint bisects the central angle, forming a right-angled triangle with one angle equal to $\frac{\pi}{3}$ and a hypotenuse equal to the radius of the circle. This perpendicular distance is calculated as $3 \cos(\frac{\pi}{3}) = \frac{3}{2}$, meaning any midpoint $(x, y)$ must satisfy the condition $\sqrt{x^2+y^2} = \frac{3}{2}$, which simplifies to the equation $x^2+y^2 = \frac{9}{4}$.

In any circle, the intersection point of any two diameters coincides with the center, and the provided area allows us to determine the radius. Solving the equations $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ simultaneously reveals the center is at $(1, -1)$. Given the area is $49\pi$, we use the formula $\pi r^2 = 49\pi$ to find a radius of $7$. Applying these coordinates and the radius to the standard circle formula $(x - 1)^2 + (y + 1)^2 = 49$ and expanding the terms leads to $x^2 - 2x + 1 + y^2 + 2y + 1 = 49$, which simplifies to $x^2 + y^2 - 2x + 2y - 47 = 0$.

The area of a circular sector is directly proportional to its central angle, so when two diameters divide a circle into sectors such that one area is thrice the other, the corresponding central angles must exist in a $1:3$ ratio. Since two intersecting diameters create supplementary angles that sum to $180^{\circ}$, we can express the angles as $\theta$ and $3\theta$, where $\theta + 3\theta = 180^{\circ}$. Solving this equation gives $4\theta = 180^{\circ}$, confirming that $\theta = 45^{\circ}$, which represents the acute angle between the two diameters.

The homogeneous quadratic equation $a x^2 + 2(a+b) x y + b y^2 = 0$ represents two lines passing through the origin that serve as these diameters. For an equation of the form $A x^2 + 2H x y + B y^2 = 0$, the tangent of the angle $\theta$ between the lines is defined by $\tan \theta = \left| \frac{2 \sqrt{H^2 - A B}}{A+B} \right|$. Substituting the values $A = a$, $H = a+b$, $B = b$, and $\theta = 45^{\circ}$ into this formula yields $1 = \frac{2 \sqrt{(a+b)^2 - a b}}{a+b}$. Squaring both sides results in $(a+b)^2 = 4((a+b)^2 - ab)$, which simplifies to $a+b = 2 \sqrt{a^2 + a b + b^2}$. Squaring once more gives $(a+b)^2 = 4(a^2 + a b + b^2)$, or $a^2 + 2ab + b^2 = 4a^2 + 4ab + 4b^2$. Rearranging these terms leads to the final relationship:
$3a^2 + 2ab + 3b^2 = 0$

Orthogonality between two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ is defined by the identity $2g_1g_2+2f_1f_2=c_1+c_2$. Applying this to a variable circle $x^2+y^2-2\alpha x-2\beta y+c=0$ and the fixed circle $x^2+y^2-p^2=0$, the orthogonality condition reduces to $0=c-p^2$, which implies $c=p^2$. The equation of the circle becomes $x^2+y^2-2\alpha x-2\beta y+p^2=0$.

Since the circle passes through $(a, b)$, the point must satisfy the equation $a^2+b^2-2\alpha a-2\beta b+p^2=0$. Replacing the center coordinates $(\alpha, \beta)$ with the variable coordinates $(x, y)$ describes the locus as $2ax+2by-(a^2+b^2+p^2)=0$.

The common chord of two intersecting circles, which passes through the points of intersection $P$ and $Q$, is determined by calculating the difference between the two circle equations. For the circles $x^2+y^2+2ax+cy+a=0$ and $x^2+y^2-3ax+dy-1=0$, subtracting the second equation from the first yields the linear equation $5ax + (c-d)y + (a+1) = 0$.

If the line $5x+by-a=0$ is indeed this common chord, its coefficients must be proportional to those of the derived linear equation, meaning $\frac{5a}{5} = \frac{c-d}{b} = \frac{a+1}{-a}$. Simplifying the first and last ratios results in the equation $a = \frac{a+1}{-a}$, which rearranges into the quadratic form $a^2 + a + 1 = 0$. Since the discriminant of this quadratic equation, $D = 1^2 - 4(1)(1)$, is equal to $-3$, no real value of $a$ satisfies the condition.

Any circle touching the x-axis with a center at $(x, y)$ possesses a radius equal to its vertical distance from the axis, which is $y$ assuming the circle lies in the upper half-plane. Because this variable circle is tangent to the fixed circle centered at $(0, 3)$ with radius $2$, the distance between the two centers must equal the sum of their radii, resulting in the equation $\sqrt{x^2 + (y-3)^2} = y + 2$. Squaring both sides gives $x^2 + (y-3)^2 = (y+2)^2$, and expanding the binomials simplifies to $x^2 + y^2 - 6y + 9 = y^2 + 4y + 4$. This rearrangement yields the equation $x^2 = 10y - 5$, which can be rewritten as $x^2 = 10(y - 0.5)$, confirming the locus describes a parabola.

The locus of a point tracing the other end of a diameter is governed by the geometric properties of the circle's center and its radius. For a circle with a diameter having endpoints at $(p, q)$ and $(x, y)$, the center is necessarily the midpoint, which is $(\frac{x+p}{2}, \frac{y+q}{2})$. Because the circle touches the $x$-axis, its radius must be equal to the absolute value of the $y$-coordinate of its center, defined as $|\frac{y+q}{2}|$. The distance from this center to the fixed point $(p, q)$ must also equal the radius, establishing the relation $(\frac{x+p}{2} - p)^2 + (\frac{y+q}{2} - q)^2 = (\frac{y+q}{2})^2$.

Simplifying the terms inside the parentheses yields $(\frac{x-p}{2})^2 + (\frac{y-q}{2})^2 = (\frac{y+q}{2})^2$. Expanding the squared terms results in $\frac{(x-p)^2}{4} + \frac{(y-q)^2}{4} = \frac{(y+q)^2}{4}$. Multiplying the entire equation by $4$ removes the denominators, giving $(x-p)^2 + (y-q)^2 = (y+q)^2$. Further expanding $(y-q)^2$ as $y^2 - 2qy + q^2$ and $(y+q)^2$ as $y^2 + 2qy + q^2$ allows for the cancellation of $y^2$ and $q^2$ from both sides. This reduction confirms that $(x-p)^2 - 2qy = 2qy$, which simplifies to $(x-p)^2 = 4qy$.

Two circles intersect orthogonally when the sum of twice the products of their corresponding center coordinates equals the sum of their constant terms. Applied to the specific case of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ intersecting $x^2 + y^2 = 4$, the condition $2g(0) + 2f(0) = c - 4$ dictates that the constant term $c$ must be $4$.

Because the variable circle passes through the point $(a, b)$, these coordinates must satisfy $a^2 + b^2 + 2ga + 2fb + c = 0$. Substituting $c = 4$ into this equation results in:
$a^2 + b^2 + 2ga + 2fb + 4 = 0$

Replacing the parameters $g$ and $f$ with the negative coordinates of the center $(-x, -y)$ transforms the expression into $a^2 + b^2 - 2ax - 2by + 4 = 0$. Rearranging the terms to isolate the variables yields the final locus equation:
$2ax + 2by - (a^2 + b^2 + 4) = 0$

A circle defined by a diameter with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ follows the equation $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$. Determining these endpoints requires finding the intersection of the line $y=x$ and the circle $x^2+y^2-2x=0$. Substituting $y=x$ into the circle equation yields $x^2+x^2-2x=0$, which simplifies to $2x^2-2x=0$. Factoring this result provides $2x(x-1)=0$, confirming the endpoints of the diameter are $(0,0)$ and $(1,1)$. Applying these points to the diameter-form equation results in $(x-0)(x-1) + (y-0)(y-1) = 0$, which simplifies to $x^2+y^2-x-y=0$.

The intersection point of any two diameters in a circle identifies its center. Solving the system $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ allows us to locate this point. Rewriting the second equation as $y = 3x - 4$ and substituting it into the first gives $2x + 3(3x - 4) = -1$. This simplifies to $2x + 9x - 12 = -1$, which leads to $11x = 11$ and therefore $x = 1$. Substituting this value back into the expression for $y$ gives $y = 3(1) - 4 = -1$, identifying the center at $(1, -1)$.

With the center known, the radius is derived from the circumference $2\pi r = 10\pi$, which simplifies to $r = 5$. Applying the standard form $(x - h)^2 + (y - k)^2 = r^2$ with center $(1, -1)$ and radius $5$ results in $(x - 1)^2 + (y + 1)^2 = 5^2$. Expanding these binomials produces $x^2 - 2x + 1 + y^2 + 2y + 1 = 25$, which simplifies to the final equation $x^2 + y^2 - 2x + 2y - 23 = 0$.

Two circles intersect at exactly two points when the distance separating their centers is less than the sum of their radii but greater than the magnitude of the difference between those radii. The first equation represents a circle centered at $(1, 3)$ with radius $r$, while the second equation, $x^2+y^2-8x+2y+8=0$, can be rearranged by completing the square into $(x-4)^2+(y+1)^2=9$, which identifies its center at $(4, -1)$ and radius as $3$. Calculating the distance between these two centers yields $\sqrt{(4-1)^2+(-1-3)^2}$, which simplifies to $\sqrt{9+16}=5$. For the circles to intersect at two distinct locations, the sum of their radii must exceed the distance between the centers, requiring $r+3 > 5$, which simplifies to $r > 2$. Furthermore, the absolute difference between the radii must remain less than the distance between the centers, expressed as $|r-3| < 5$. Solving this inequality leads to $-5 < r-3 < 5$, which, when constrained by the requirement for a positive radius, results in $r < 8$. Combining these conditions establishes that the circles intersect at two distinct points only when $2 < r < 8$.

The center of a circle is uniquely defined by the intersection point of its diameters, while the radius is constrained by the circle's total area. Solving the linear system $2x - 3y = 5$ and $3x - 4y = 7$ reveals that the center is located at $x = 1$ and $y = -1$. Since the area is given as 154 sq. units, we apply the relation $\pi r^2 = 154$ to determine that $r^2 = 49$. Inserting these values into the standard circle equation form $(x - h)^2 + (y - k)^2 = r^2$ yields the expansion $x^2 - 2x + 1 + y^2 + 2y + 1 = 49$, which simplifies to $x^2 + y^2 - 2x + 2y = 47$.

Since the circle passes through $(0,0)$ and $(1,0)$, its center must lie on the perpendicular bisector of the line segment joining these points, which is the vertical line $x = \frac{1}{2}$. If we denote the center as $(\frac{1}{2}, k)$, the radius $r$ is the distance from this point to the origin, which is given by $\sqrt{(\frac{1}{2})^2 + k^2}$.

The circle $x^2+y^2=9$ is centered at the origin with a radius of 3. Because both $(0,0)$ and $(1,0)$ lie inside this larger circle, the two circles touch internally. In such a configuration, the distance between the centers is equal to the difference between the radius of the larger circle and the radius of the smaller circle, resulting in the equation $r = 3 - r$.

Solving this gives $2r = 3$, so $r = \frac{3}{2}$. Setting the distance formula equal to this radius, $\sqrt{(\frac{1}{2})^2 + k^2} = \frac{3}{2}$, and squaring both sides produces $\frac{1}{4} + k^2 = \frac{9}{4}$. Subtracting $\frac{1}{4}$ from both sides leaves $k^2 = 2$, which means $k = \pm \sqrt{2}$. The possible centers are therefore $(\frac{1}{2}, \sqrt{2})$ and $(\frac{1}{2}, -\sqrt{2})$.

The set of points occupied by circles of radius 3 whose centers lie on the circle $x^2+y^2=25$ forms an annular region centered at the origin. Because the centers are fixed at a distance of 5 from the origin, any point $P$ within one of these circles has a distance $OP$ that is bounded by the distance from the origin to the circle's center plus or minus the circle's radius. This implies that the distance $OP$ must fall between $5-3=2$ and $5+3=8$. By squaring these boundary values to represent the distance from the origin as $x^2+y^2$, we arrive at the locus $4 \leq x^2+y^2 \leq 64$.

The homogenization of the circle equation $x^2 + y^2 = 1$ using the line equation $y = mx + 1$ helps represent the pair of lines connecting the origin to the chord's intersection points. Since the line equation can be written as $y - mx = 1$, substituting the square of this expression into the circle equation gives $x^2 + y^2 = (y - mx)^2$. Expanding this yields $x^2 + y^2 = y^2 - 2mxy + m^2x^2$, which simplifies to $x^2(1 - m^2) + 2mxy = 0$.

This quadratic equation represents the pair of lines, and the angle $\theta$ between them is determined by the formula $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$. Here, the coefficients are $a = 1 - m^2$, $h = m$, and $b = 0$. Given $\theta = 45^{\circ}$, we have $\tan 45^{\circ} = 1$, which leads to the equation $1 = \left| \frac{2\sqrt{m^2 - 0}}{1 - m^2} \right|$, further simplifying to $\left| \frac{2m}{1 - m^2} \right| = 1$.

Setting $1 - m^2 = \pm 2m$ results in the quadratic equation $m^2 \pm 2m - 1 = 0$. Applying the quadratic formula $m = \frac{\mp 2 \pm \sqrt{4 - 4(1)(-1)}}{2}$ simplifies to $m = \frac{-2 \pm 2\sqrt{2}}{2}$, providing the value $m = -1 \pm \sqrt{2}$.

The centroid of an equilateral triangle is equidistant from its three vertices, which places it at the center of the circle that passes through those vertices. In an equilateral triangle, the median serves as the altitude, and the centroid divides this segment into two parts with a 2:1 length ratio, where the longer segment represents the distance from the centroid to any vertex. Given that the entire median measures $3a$, the radius of the circle is simply two-thirds of that length, resulting in a radius of $2a$. With the center at the origin, the equation of the circle follows the form $x^2+y^2=r^2$, and substituting $2a$ for the radius yields $x^2+y^2=4a^2$.

A circle's center is the point of intersection for its normal lines, and solving the equations $y+2x=\sqrt{11}+7\sqrt{7}$ and $2y+x=2\sqrt{11}+6\sqrt{7}$ simultaneously provides the coordinates for the center $(h, k) = \left(\frac{8\sqrt{7}}{3}, \sqrt{11}+\frac{5\sqrt{7}}{3}\right)$. Because the radius $r$ represents the perpendicular distance from the center to the tangent line $\sqrt{11}y-3x=\frac{5\sqrt{77}}{3}+11$, we use the distance formula to determine:
$r = \frac{\left|-3\left(\frac{8\sqrt{7}}{3}\right) + \sqrt{11}\left(\sqrt{11}+\frac{5\sqrt{7}}{3}\right) - \left(\frac{5\sqrt{77}}{3}+11\right)\right|}{\sqrt{(-3)^2+(\sqrt{11})^2}}$
This simplifies to $r = \frac{|-8\sqrt{7}+11+\frac{5\sqrt{77}}{3}-\frac{5\sqrt{77}}{3}-11|}{\sqrt{20}}$, which evaluates to $r = \frac{8\sqrt{7}}{2\sqrt{5}} = 4\sqrt{\frac{7}{5}}$. With the center and radius established, calculating the expression $(5h-8k)^2+5r^2$ begins by finding $5h-8k = 5\left(\frac{8\sqrt{7}}{3}\right) - 8\left(\sqrt{11}+\frac{5\sqrt{7}}{3}\right) = -8\sqrt{11}$. Squaring this value gives $(-8\sqrt{11})^2 = 704$, and the squared radius component $5r^2$ becomes $5\left(16 \cdot \frac{7}{5}\right) = 112$. Adding these results together yields the final value of 816.

To determine the geometry of the circle defined by $x^2 + y^2 + 2gx + 2fy + 25 = 0$, recognize that the center is situated at $(-g, -f)$ and the radius squared is given by $r^2 = g^2 + f^2 - 25$. Since the area of an equilateral triangle inscribed within a circle is calculated as $\frac{3\sqrt{3}}{4} r^2$, equating this expression to $27\sqrt{3}$ confirms that $r^2 = 36$. Because the center lies on the line $2x - y = 4$, substituting the coordinates $-g$ and $-f$ yields the linear constraint $-2g + f = 4$, which simplifies to $f = 2g + 4$.

Combining these relationships into the radius formula $g^2 + f^2 = r^2 + 25$ leads to the following equation:

$g^2 + (2g + 4)^2 = 36 + 25 = 61$

Expanding and solving the quadratic equation $5g^2 + 16g - 45 = 0$ provides two possible pairs for the center. Selecting the values that place the center $(-g, -f)$ in the first quadrant identifies the coordinates as $(5, 6)$. The perpendicular distance $d$ from this center to the chord line $x = 1$ is $|5 - 1| = 4$. Using the relationship for chord length $L = 2\sqrt{r^2 - d^2}$, the square of the length is $4(r^2 - d^2)$, which calculates to $4(36 - 16) = 80$.