Circles – JEE Main MathematicsPractice Questions & PYQs

The radius squared of a circle containing a chord is given by the sum of the square of the perpendicular distance from the circle's center to the chord and the square of the chord's half-length. For the first circle $x^2+y^2-2\sqrt{2}x-6\sqrt{2}y+14=0$, we identify the center $C_1$ as $(\sqrt{2}, 3\sqrt{2})$ and calculate the radius squared $R_1^2 = (\sqrt{2})^2 + (3\sqrt{2})^2 - 14 = 2 + 18 - 14 = 6$. Since the chord is a diameter of this circle, its half-length is simply the radius of the first circle, $\sqrt{6}$, so the square of the half-length is $6$.

The second circle $(x-2\sqrt{2})^2+(y-2\sqrt{2})^2=r^2$ has its center at $C_2 = (2\sqrt{2}, 2\sqrt{2})$. For the diameter of the first circle to serve as a chord of the second, it must be oriented such that its endpoints are equidistant from $C_2$, which geometrically requires the diameter to be perpendicular to the line connecting $C_1$ and $C_2$. Therefore, the perpendicular distance $d$ from $C_2$ to the chord is equal to the distance between the centers $C_1$ and $C_2$. Calculating this distance squared, we get $d^2 = (\sqrt{2}-2\sqrt{2})^2 + (3\sqrt{2}-2\sqrt{2})^2 = (-\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$. Substituting these values into the geometric relationship $r^2 = d^2 + R_1^2$ yields $r^2 = 4 + 6 = 10$.

A line $y=mx+c$ is tangent to a circle if the perpendicular distance from the center to the line equals the radius $r$. Rewriting the line equation as $mx-y+c=0$, the distance from a center $(x_0, y_0)$ to the line is expressed as $\frac{|mx_0-y_0+c|}{\sqrt{m^2+1}}$. For circles $C_1$ with center $(0,0)$, $C_2$ with center $(1,1)$, and $C_3$ with center $(2,1)$, these distances are $\frac{|c|}{\sqrt{m^2+1}}$, $\frac{|m-1+c|}{\sqrt{m^2+1}}$, and $\frac{|2m-1+c|}{\sqrt{m^2+1}}$, all of which must equal $r$. Because $C_1$ and $C_3$ lie on one side of the line and $C_2$ lies on the other, the signed distances for $C_1$ and $C_3$ share the same sign, while the signed distance for $C_2$ must have the opposite sign.

Equating the magnitudes to satisfy this geometric condition requires solving the system $c = -(m-1+c)$ and $m-1+c = -(2m-1+c)$, which simplifies to the linear equations $m+2c=1$ and $3m+2c=2$. Solving this system, $2m=1$ yields $m=1/2$, which subsequently gives $c=1/4$. Substituting these values into the distance formula for $C_1$ produces the radius $r = \frac{|1/4|}{\sqrt{(1/2)^2+1}} = \frac{1/4}{\sqrt{5}/2} = \frac{1}{2\sqrt{5}}$, leading to $r^2 = 1/20$. The requested value $20(r^2+c)$ is therefore calculated as $20(1/20 + 1/4) = 20(1/20 + 5/20) = 20(6/20) = 6$.

The lines $L_1: 4x-3y+K_1=0$ and $L_2: 4x-3y+K_2=0$ are parallel because they share the same slope of $4/3$. Since the circle is tangent to both, the segment connecting the points of tangency lies along a line passing through the center that is perpendicular to these parallel tangents; therefore, the center of the circle is the midpoint of the segment joining $(-1, 2)$ and $(3, -6)$, which evaluates to $(1, -2)$.

Substituting the known points of tangency into the equations of the lines allows us to identify the constants $K_1$ and $K_2$. Inserting $(-1, 2)$ into $4x-3y+K_1=0$ yields $K_1=10$, while substituting $(3, -6)$ into $4x-3y+K_2=0$ results in $K_2=-30$. The diameter of the circle corresponds to the perpendicular distance between these parallel lines, calculated using the distance formula $d = \frac{|K_1-K_2|}{\sqrt{A^2+B^2}}$, which yields $d = 8$. With a diameter of 8, the radius is 4. Applying the standard form $(x-h)^2+(y-k)^2=r^2$ with center $(1, -2)$ and radius $r=4$ leads to the final equation:

$(x-1)^2+(y+2)^2=16$

The first circle $x^{2}+y^{2}+6x+8y+16=0$ can be rewritten in the standard form $(x+3)^{2}+(y+4)^{2}=3^{2}$, which identifies its center $O_1$ as $(-3, -4)$ and its radius $r_1=3$. Completing the square for the second circle $x^{2}+y^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y=k+6\sqrt{3}+8\sqrt{6}$ yields $(x+3-\sqrt{3})^{2}+(y+4-\sqrt{6})^{2}=k+34$, indicating a center $O_2$ at $(\sqrt{3}-3, \sqrt{6}-4)$ and a radius $r_2=\sqrt{k+34}$. For these two circles to touch internally, the distance between their centers must equal the absolute difference of their radii, $|r_2-r_1|$. The distance $d$ between $O_1$ and $O_2$ is calculated as $\sqrt{(\sqrt{3}-3 - (-3))^2 + (\sqrt{6}-4 - (-4))^2} = \sqrt{(\sqrt{3})^2 + (\sqrt{6})^2} = 3$. Setting this equal to $|r_2-r_1| = \sqrt{k+34}-3$ results in $\sqrt{k+34}=6$, which implies $k=2$ and confirms the radius of the second circle is $r_2=6$.

Because the circles touch internally, the point of contact $P(\alpha, \beta)$ lies on the line passing through the centers $O_1$ and $O_2$, positioned such that the smaller circle is contained within the larger one. Starting from $O_1(-3, -4)$, we move in the direction opposite to the vector $\vec{O_1O_2} = (\sqrt{3}, \sqrt{6})$ by a distance equal to the smaller radius $r_1=3$. The unit vector in the direction of $O_1O_2$ is $\frac{1}{3}(\sqrt{3}, \sqrt{6})$, so the contact point $P$ is located at $O_1 - 3 \times \frac{1}{3}(\sqrt{3}, \sqrt{6}) = (-3-\sqrt{3}, -4-\sqrt{6})$. Consequently, $\alpha+\sqrt{3} = -3$ and $\beta+\sqrt{6} = -4$, and the final expression $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ evaluates to $(-3)^{2}+(-4)^{2} = 25$.

The core principle for a circle to lie entirely within the fourth quadrant is that its radius must be sufficiently small such that the distance from the center to both coordinate axes remains greater than the radius, ensuring no part of the circle crosses the axes into other quadrants. For the given circle equation $4x^2+4y^2-12x+8y+k=0$, we first rewrite it in standard form as $x^2+y^2-3x+2y+\frac{k}{4}=0$, which identifies the center at $(\frac{3}{2}, -1)$ and establishes the squared radius as $r^2 = (\frac{3}{2})^2 + (-1)^2 - \frac{k}{4} = \frac{13}{4} - \frac{k}{4}$. Since the distance from the center to the $x$-axis is $1$ and the distance to the $y$-axis is $\frac{3}{2}$, the condition for the circle to be confined to the fourth quadrant is $r < 1$, which implies $r^2 < 1$. Substituting the expression for $r^2$ results in $\frac{13}{4} - \frac{k}{4} < 1$, which simplifies to $k > 9$.

Additionally, the requirement for the point $(1, -\frac{1}{3})$ to lie on or inside the circle necessitates that substituting these coordinates into the circle equation yields a non-positive value. Substituting the point into the original equation gives $4(1)^2 + 4(-\frac{1}{3})^2 - 12(1) + 8(-\frac{1}{3}) + k \le 0$, which simplifies to $4 + \frac{4}{9} - 12 - \frac{8}{3} + k \le 0$. Combining the constant terms leads to $-8 - \frac{20}{9} + k \le 0$, resulting in $k \le \frac{92}{9}$. Therefore, the circle satisfies both conditions for all values of $k$ in the interval $(9, \frac{92}{9}]$.

Any circle defined by $x^2+y^2-2gx+6y-19c=0$ has its center located at $(g, -3)$, and since this circle passes through the point $(6, 1)$, we can substitute these coordinates to obtain $36+1-12g+6-19c=0$, which simplifies to $12g+19c=43$. Given that the center $(g, -3)$ also satisfies the linear equation $x-2cy=8$, substituting these coordinates results in $g-2c(-3)=8$, or $g+6c=8$, allowing us to solve the system to find $g=2$ and $c=1$. With the circle's equation established as $x^2+y^2-4x+6y-19=0$, we determine the length of the intercept on the $x$-axis by finding the difference between the roots of the quadratic equation obtained by setting $y=0$, which is $x^2-4x-19=0$, resulting in a length of $2\sqrt{23}$.

The equation of a circle having the endpoints of its diameter at $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$. This can be expanded to $x^2+y^2-(x_1+x_2)x-(y_1+y_2)y + (x_1x_2+y_1y_2) = 0$. According to Vieta's formulas, the sum and product of the roots for a quadratic equation $t^2+pt+q=0$ are $-p$ and $q$, respectively. For the abscissae $x_1$ and $x_2$ originating from $x^2-4x-6=0$, the sum $x_1+x_2$ is $4$ and the product $x_1x_2$ is $-6$. Similarly, for the ordinates $y_1$ and $y_2$ from $y^2+2y-7=0$, the sum $y_1+y_2$ is $-2$ and the product $y_1y_2$ is $-7$.

Substituting these values into the expanded form of the circle equation yields $x^2+y^2-4x-(-2)y + (-6-7) = 0$, which simplifies to $x^2+y^2-4x+2y-13=0$. Comparing this result with the given equation $x^2+y^2+2ax+2by+c=0$, we determine that $2a=-4$, $2b=2$, and $c=-13$. These relations establish that $a=-2$, $b=1$, and $c=-13$. Computing the required expression gives $a+b-c = -2+1-(-13)$, which simplifies to $12$.

The number of integer coordinate pairs in the intersection $S \cap T$ arises from evaluating the constraints of both the ellipse and the circle. The ellipse region $S$ is restricted to natural numbers and follows the inequality $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1$, which constrains $x$ to the range $1 \leq x \leq 7$. Possible values for $y$ exist at each integer $x$, where only those satisfying the circle condition $(x-7)^2 + (y-4)^2 \leq 36$ are counted.

For $x=1$, the ellipse condition allows $y \in \{2, 3, 4, 5, 6\}$. Substituting these into the circle inequality $(1-7)^2 + (y-4)^2 \leq 36$ simplifies to $36 + (y-4)^2 \leq 36$, which only holds for $y=4$. For $x=2$, the ellipse allows $y \in \{2, 3, 4, 5, 6\}$, and checking the circle $(2-7)^2 + (y-4)^2 = 25 + (y-4)^2 \leq 36$ shows that all five points satisfy the inequality. For $x=3$, the ellipse allows $y \in \{1, 2, 3, 4, 5, 6, 7\}$, and the circle $(3-7)^2 + (y-4)^2 = 16 + (y-4)^2 \leq 36$ is satisfied by all seven points since $(y-4)^2 \leq 20$ for all $y \in \{1, 2, 3, 4, 5, 6, 7\}$.

For $x=4$, the ellipse allows $y \in \{2, 3, 4, 5, 6\}$, all of which satisfy the circle inequality $9 + (y-4)^2 \leq 36$. For $x=5$, the ellipse permits $y \in \{2, 3, 4, 5, 6\}$, all of which satisfy the circle inequality $4 + (y-4)^2 \leq 36$. For $x=6$, the ellipse permits $y \in \{3, 4, 5\}$, all of which satisfy the circle inequality $1 + (y-4)^2 \leq 36$. Finally, at $x=7$, the ellipse only allows $y=4$, which satisfies the circle inequality $0 + (4-4)^2 \leq 36$. The total count of intersection points is $1 + 5 + 7 + 5 + 5 + 3 + 1 = 27$.

The center of any circle passing through two fixed points must lie on the perpendicular bisector of the line segment joining those points. For the points $A(2,-1)$ and $B(3,4)$, the midpoint of the chord is $(\frac{5}{2}, \frac{3}{2})$ and the slope of the chord is $\frac{4 - (-1)}{3 - 2} = 5$. Since the perpendicular bisector must be perpendicular to the chord, it has a slope of $-\frac{1}{5}$, leading to the equation:
$y - \frac{3}{2} = -\frac{1}{5}\left(x - \frac{5}{2}\right) \implies x + 5y = 10$

Because the center also lies on the given circle $(x-5)^2+(y-1)^2=\frac{13}{2}$, we can substitute $x = 10 - 5y$ into this equation to find the coordinates of the center. This substitution gives $(10 - 5y - 5)^2 + (y - 1)^2 = \frac{13}{2}$, which simplifies to $25(1-y)^2 + (y-1)^2 = \frac{13}{2}$. Solving $26(y-1)^2 = \frac{13}{2}$ yields $(y-1)^2 = \frac{1}{4}$, which implies $y = 1 \pm \frac{1}{2}$. We reject the case $y = \frac{3}{2}$ because it corresponds to $x = \frac{5}{2}$, which is the midpoint of the chord $AB$; this would mean $AB$ is a diameter, which is explicitly forbidden by the problem statement. Using $y = \frac{1}{2}$ instead provides the center coordinates $(\frac{15}{2}, \frac{1}{2})$.

The squared radius $r^2$ is the squared distance between the center $(\frac{15}{2}, \frac{1}{2})$ and point $A(2, -1)$, which is determined by the distance formula:
$r^2 = \left(\frac{15}{2} - 2\right)^2 + \left(\frac{1}{2} - (-1)\right)^2 = \left(\frac{11}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}$

The tangent to circle $C_1$ at $M(-1, 1)$ is determined by the equation $x x_1 + y y_1 = r^2$, which simplifies to $-x + y = 2$, or $y = x + 2$. Substituting this relationship into the equation for circle $C_2$, $(x-3)^2 + (y-2)^2 = 5$, leads to the quadratic $x^2 - 3x + 2 = 0$, identifying the intersection points at $A(1, 3)$ and $B(2, 4)$. Applying the tangent equation $T=0$ for circle $C_2$ at these points results in the linear equations $2x - y + 1 = 0$ for point $A$ and $x - 2y + 6 = 0$ for point $B$, which intersect at the vertex $N(\frac{4}{3}, \frac{11}{3})$.

Calculating the area of triangle $ANB$ with vertices $A(1, 3)$, $B(2, 4)$, and $N(\frac{4}{3}, \frac{11}{3})$ uses the coordinate geometry area formula $\frac{1}{2} |x_A(y_B - y_N) + x_B(y_N - y_A) + x_N(y_A - y_B)|$. Substituting the coordinate values yields $\frac{1}{2} |1(4 - \frac{11}{3}) + 2(\frac{11}{3} - 3) + \frac{4}{3}(3 - 4)|$, which simplifies to $\frac{1}{2} | \frac{1}{3} + \frac{4}{3} - \frac{4}{3} |$, resulting in an area of $\frac{1}{6}$.

The area of triangle $PQR$ is determined by the length of its base, which is the chord passing through the center, and its altitude, which is the perpendicular distance from point $P$ to that chord. By completing the square for the circle equation $x^2+y^2-x+2 y=\frac{11}{4}$, the standard form becomes $(x-\frac{1}{2})^2+(y+1)^2=4$, indicating the center is $C(\frac{1}{2},-1)$ and the radius $r=2$. Since the line containing $Q$ and $R$ passes through the center, $QR$ is a diameter with length $QR=4$.

The position of point $P$ on the circle creates a radius $CP$ of length $2$. Given that the chord $QR$ makes an angle of $\frac{\pi}{4}$ with this radius, the perpendicular height $h$ from $P$ to $QR$ is defined by $h=r\sin(\frac{\pi}{4})$. Substituting the values gives $h=2\sin(\frac{\pi}{4})=\sqrt{2}$. The area of the triangle is then calculated as $\text{Area} = \frac{1}{2} \times 4 \times \sqrt{2} = 2\sqrt{2}$ square units.

The geometric property that the radius of a circle is perpendicular to the tangent line at the point of contact implies that the segment connecting the center $P$ to the point of tangency $Q$ must be perpendicular to the tangent line $L_2$. Since the line $L_1: 4x+3y+2=0$ contains the center $P$ and intersects $L_2: 3x-4y-11=0$ at the point $Q$, the length of the segment $PQ$ is equivalent to the radius of the circle, which is 5.

The coordinates of $Q$ are found by solving the system of equations defined by $L_1$ and $L_2$, which yields the intersection point $Q(1, -2)$. Given that the center $P(x, y)$ must lie on $L_1$, its coordinates satisfy $4x+3y+2=0$, or $y = -\frac{4x+2}{3}$. Using the distance formula for $PQ=5$ gives the equation $(x-1)^2 + (y+2)^2 = 25$. Substituting the expression for $y$ into this equation leads to $(x-1)^2 + (-\frac{4x+2}{3} + 2)^2 = 25$, which simplifies to $(x-1)^2(1 + \frac{16}{9}) = 25$, or $(x-1)^2 = 9$. This results in $x$ values of $4$ and $-2$. Because the circle must reside completely below the $x$-axis, the only valid center is $P(4, -6)$, as the alternative coordinates would place part of the circle above the $x$-axis.

The perpendicular distance from the center $P(4, -6)$ to the line $5x-12y+51=0$ is determined by applying the standard distance formula:
$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$
Substituting the coordinates of $P$ and the coefficients of the line yields:
$d = \frac{|5(4) - 12(-6) + 51|}{\sqrt{5^2 + (-12)^2}} = \frac{|20 + 72 + 51|}{\sqrt{169}} = \frac{143}{13}$
This calculation results in a final distance of 11.

Finding the intersection point $Q$ of the tangents drawn at $O(0,0)$ and $P(1+\sqrt{5}, 2)$ to the circle $x^2+y^2-2x-4y=0$ is the core principle for calculating the area of triangle $OPQ$. The tangent equation to a circle $x^2+y^2+2gx+2fy+c=0$ at a point $(x_1, y_1)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. For this specific circle, where $g=-1$ and $f=-2$, the equation becomes $xx_1 + yy_1 - (x+x_1) - 2(y+y_1) = 0$.

For the point $O(0,0)$, this formula results in the equation $x+2y=0$. Applying the same formula for $P(1+\sqrt{5}, 2)$, the expression becomes $x(1+\sqrt{5}) + 2y - (x+1+\sqrt{5}) - 2(y+2) = 0$. Simplifying this leads to $x\sqrt{5} = 5+\sqrt{5}$, which yields $x = 1+\sqrt{5}$. Substituting this value into the equation $x+2y=0$ provides the coordinate $y = -\frac{1+\sqrt{5}}{2}$. Thus, the intersection point $Q$ is $(1+\sqrt{5}, -\frac{1+\sqrt{5}}{2})$.

The triangle $OPQ$ has vertices $O(0,0)$, $P(1+\sqrt{5}, 2)$, and $Q(1+\sqrt{5}, -\frac{1+\sqrt{5}}{2})$. Since $P$ and $Q$ share the same $x$-coordinate, the side $PQ$ is vertical, and its length represents the base of the triangle. This length is the absolute difference of the $y$-coordinates: $|2 - (-\frac{1+\sqrt{5}}{2})| = \frac{5+\sqrt{5}}{2}$. The height of the triangle from vertex $O$ to this vertical base is the horizontal distance, which is $1+\sqrt{5}$. The area is computed as:

$\text{Area} = \frac{1}{2} \times \left(\frac{5+\sqrt{5}}{2}\right) \times (1+\sqrt{5})$

$\text{Area} = \frac{1}{4} (5 + 5\sqrt{5} + \sqrt{5} + 5) = \frac{10+6\sqrt{5}}{4} = \frac{5+3\sqrt{5}}{2}$

In any equilateral triangle, the orthocentre coincides perfectly with the centroid. The coordinates $(h, k)$ of the centroid are found by averaging the vertices $A(\sin t, -\cos t)$, $B(\cos t, \sin t)$, and $C(a, b)$, yielding $h = \frac{\sin t + \cos t + a}{3}$ and $k = \frac{-\cos t + \sin t + b}{3}$. Isolating the trigonometric terms gives $3h - a = \sin t + \cos t$ and $3k - b = -\cos t + \sin t$. Squaring and adding both equations eliminates the trigonometric dependency using the identity $(\sin t + \cos t)^2 + (-\cos t + \sin t)^2 = 2(\sin^2 t + \cos^2 t) = 2$, which results \in $(3h - a)^2 + (3k - b)^2 = 2$.

Dividing this equation by 9 produces the standard form of a circle, $\left(h - \frac{a}{3}\right)^2 + \left(k - \frac{b}{3}\right)^2 = \frac{2}{9}$, which identifies the centre of the locus as $\left(\frac{a}{3}, \frac{b}{3}\right)$. Since the problem specifies that the orthocentre lies on a circle with centre $\left(1, \frac{1}{3}\right)$, equating the coordinates gives $a = 3$ and $b = 1$. Applying these values to the required expression results \in $a^2 - b^2 = 3^2 - 1^2 = 8$.

The circle with $PQ$ as a diameter is defined by the equation $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$, where $(x_1, y_1)$ and $(x_2, y_2)$ represent the coordinates of points $P$ and $Q$. Expanding this yields $x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$. Based on Vieta's formulas for the quadratic $2x^2 - rx + p = 0$, we have $x_1 + x_2 = \frac{r}{2}$ and $x_1x_2 = \frac{p}{2}$, while the quadratic $x^2 - sx - q = 0$ provides $y_1 + y_2 = s$ and $y_1y_2 = -q$. Substituting these relations into the expanded circle equation results in $x^2 - \frac{r}{2}x + \frac{p}{2} + y^2 - sy - q = 0$, which can be rewritten as $2x^2 + 2y^2 - rx - 2sy + p - 2q = 0$ after scaling by two. Comparing this expression to the provided circle equation $2(x^2 + y^2) - 11x - 14y - 22 = 0$, the coefficients correspond to $r = 11$, $2s = 14$ which gives $s = 7$, and the constant part $p - 2q = -22$. Combining these results to evaluate $2r + s - 2q + p$ yields $2(11) + 7 - 22 = 7$.