Circles – JEE Main MathematicsPractice Questions & PYQs

The position of a point dividing a line segment in a specific ratio is determined by the section formula, where the coordinates of the moving point $P(h, k)$ depend on the fixed point $A(1, 2)$ and a variable point $B$ on the circle $x^2 + y^2 = 16$. Parameterizing $B$ as $(4\cos\theta, 4\sin\theta)$ lets us express $P$ as $(\frac{12\cos\theta + 2}{5}, \frac{12\sin\theta + 4}{5})$, which yields the relations $12\cos\theta = 5h - 2$ and $12\sin\theta = 5k - 4$. Squaring and adding these equations eliminates $\theta$ using the identity $\cos^2\theta + \sin^2\theta = 1$, resulting in the locus $(5h - 2)^2 + (5k - 4)^2 = 144$. By rearranging this into the standard form $(h - \frac{2}{5})^2 + (k - \frac{4}{5})^2 = \frac{144}{25}$, we identify the center of the locus as $C(\frac{2}{5}, \frac{4}{5})$. The length of the line segment $AC$ is calculated as $\sqrt{(1 - \frac{2}{5})^2 + (2 - \frac{4}{5})^2} = \sqrt{(\frac{3}{5})^2 + (\frac{6}{5})^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \frac{\sqrt{45}}{5} = \frac{3\sqrt{5}}{5}$.

The circle in the first quadrant touching both axes has center $(r, r)$ and radius $r$, satisfying the equation $(x-r)^2 + (y-r)^2 = r^2$. Because the point $P(\alpha, \beta)$ lies on the circle, we expand the equation to obtain $\alpha^2 + \beta^2 - 2r(\alpha + \beta) + r^2 = 0$.

The points of tangency $A(r, 0)$ and $B(0, r)$ define the line $x + y = r$, and the perpendicular distance $PQ$ from $P(\alpha, \beta)$ to this line is given by $PQ = \frac{|\alpha + \beta - r|}{\sqrt{1^2 + 1^2}}$. Given $PQ = 11$, squaring both sides yields $121 = \frac{(\alpha + \beta - r)^2}{2}$, which simplifies to $242 = (\alpha + \beta - r)^2$.

Expanding the expression $(\alpha + \beta - r)^2$ gives $\alpha^2 + \beta^2 + r^2 + 2\alpha\beta - 2r(\alpha + \beta)$, and substituting the circle identity $\alpha^2 + \beta^2 - 2r(\alpha + \beta) + r^2 = 0$ into this expansion allows the terms to cancel out. This process leaves $242 = 2\alpha\beta$, confirming that $\alpha\beta = 121$.

Understanding the transformation of geometric shapes relies on identifying their fundamental parameters and applying coordinate geometry to reflect those changes. The circle given by $x^2+y^2-4 x-6 y+11=0$ has its center at $(2,3)$ and a radius of $\sqrt{2}$. The equation of the tangent line at the point $(3,2)$ is derived as $x - y - 1 = 0$, which has a slope of $1$, implying it makes a $45^\circ$ angle with the $x$-axis. Moving the circle upwards 4 units along this tangent corresponds to translating the center from $(2,3)$ by a vector of $4(\cos 45^\circ, \sin 45^\circ)$, which equals $(2\sqrt{2}, 2\sqrt{2})$. Consequently, the new center $A$ is located at $(2+2\sqrt{2}, 3+2\sqrt{2})$.

The circle $C_2$ is defined as the image of $C_1$ reflected in the tangent line $x - y - 1 = 0$. By applying the reflection formula for a point across a line, the new center $B$ is determined using the relation $\frac{x_B - x_A}{1} = \frac{y_B - y_A}{-1} = -2\frac{x_A - y_A - 1}{1^2 + (-1)^2}$. Substituting the coordinates of $A$, this simplifies to $\frac{x_B - (2+2\sqrt{2})}{1} = \frac{y_B - (3+2\sqrt{2})}{-1} = 2$, which yields $x_B = 4+2\sqrt{2}$ and $y_B = 1+2\sqrt{2}$. The feet of the perpendiculars from $A$ and $B$ to the $x$-axis, denoted as $M$ and $N$, are $(2+2\sqrt{2}, 0)$ and $(4+2\sqrt{2}, 0)$, respectively. These points define a trapezium $AMNB$ with parallel vertical sides $AM = 3+2\sqrt{2}$ and $BN = 1+2\sqrt{2}$, separated by a horizontal distance $MN = 2$. The area of the trapezium is thus calculated as $\frac{1}{2}(AM + BN) \times MN = \frac{1}{2}(4+4\sqrt{2}) \times 2 = 4(1+\sqrt{2})$.

Because the line segment $AB$ has length $\lambda$ and both endpoints lie on a circle with radius $\lambda$, the triangle formed by the circle's center $O$ and the chord $AB$ is equilateral. This geometric property establishes that the distance from the center $O$ to vertex $A$ is exactly $\lambda$ and the interior angle at vertex $A$ is $60^\circ$. The point $P$, which divides $AB$ in the ratio $2:3$, is positioned at a distance of $\frac{2}{5}\lambda$ from point $A$.

The distance of $P$ from the center $O$ remains constant throughout the movement of the chord, effectively defining the radius of its circular locus. Applying the law of cosines within triangle $OAP$ yields the relationship $OP^2 = OA^2 + AP^2 - 2(OA)(AP)\cos(60^\circ)$. Substituting the known lengths results in $OP^2 = \lambda^2 + (\frac{2}{5}\lambda)^2 - 2(\lambda)(\frac{2}{5}\lambda)(\frac{1}{2})$. This simplifies to $OP^2 = \lambda^2 + \frac{4}{25}\lambda^2 - \frac{2}{5}\lambda^2$, which calculates to $OP^2 = \frac{19}{25}\lambda^2$. Taking the square root, the radius of this locus is $\frac{\sqrt{19}}{5}\lambda$.

The center of a circle $(h, k)$ must be equidistant from any two tangent lines, which implies that the center lies on the angle bisectors of the lines defining the triangle. Given the three tangent lines, the center of the circle functions as either the incenter or one of the excenters of the triangle formed by these lines, meaning it must satisfy the angle bisector equations.

Considering the lines $4x-3y+1=0$ and $3x-4y+6=0$, the angle bisectors are determined by equating the normalized distances to each line as follows:
$\frac{4x-3y+1}{5} = \pm \frac{3x-4y+6}{5}$
Taking the positive case, we obtain $4x-3y+1 = 3x-4y+6$. Simplifying this expression by isolating the variables on one side leads to the equation $x+y=5$. Since the center $(h, k)$ of the circle must lie on this bisector, substituting these coordinates yields $h+k=5$.

The problem centers on the concept of the chord of contact, which is the line connecting the points of tangency when tangents are drawn from an external point $S(\alpha, \beta)$ to a circle. First, determine the equation of the circle with center $(2,3)$ and radius 4, which expands to $(x-2)^2 + (y-3)^2 = 16$, or simplified, $x^2 + y^2 - 4x - 6y - 3 = 0$. The general equation for the chord of contact from a point $(\alpha, \beta)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $x\alpha + y\beta + g(x+\alpha) + f(y+\beta) + c = 0$. Substituting the values for our specific circle, we get $x\alpha + y\beta - 2(x+\alpha) - 3(y+\beta) - 3 = 0$, which simplifies to $x(\alpha - 2) + y(\beta - 3) - (2\alpha + 3\beta + 3) = 0$.

Since this line is identical to the given chord equation $x+y-3=0$, the coefficients of these two linear equations must be proportional. Comparing the coefficients gives the ratios $\frac{\alpha - 2}{1} = \frac{\beta - 3}{1} = \frac{2\alpha + 3\beta + 3}{3}$. Equating the first two parts yields $\alpha - 2 = \beta - 3$, or $\alpha - \beta = -1$. Equating the first and third parts leads to $3\alpha - 6 = 2\alpha + 3\beta + 3$, which simplifies to $\alpha - 3\beta = 9$. Solving this system of linear equations by substituting $\alpha = \beta - 1$ into the second equation gives $-2\beta = 10$, resulting in $\beta = -5$ and consequently $\alpha = -6$. Finally, calculating the required value $4\alpha - 7\beta$ involves substituting these coordinates to obtain $4(-6) - 7(-5) = -24 + 35$, which equals 11.

Since reflection is an isometry, it preserves the radius of any geometric figure; therefore, the radius of the original circle $C_1$ is identical to the radius of the reflected circle $C_2$. The given equation $x^2+y^2-4x-2y+5-\alpha=0$ corresponds to a circle with center $(2, 1)$ and radius $r = \sqrt{2^2+1^2-(5-\alpha)} = \sqrt{\alpha}$. We determine the center $(f, g)$ of the reflected circle by calculating the mirror image of $(2, 1)$ about the line $2x-y+1=0$ using the formula $\frac{x-2}{2} = \frac{y-1}{-1} = -2 \frac{2(2)-1+1}{2^2+(-1)^2}$. Solving this yields $\frac{x-2}{2} = \frac{y-1}{-1} = -\frac{8}{5}$, so the new center is at $f = -\frac{6}{5}$ and $g = \frac{13}{5}$. By rewriting the equation of $C_2$ in the standard form $x^2+y^2-2fx-2gy+\frac{36}{5}=0$, we identify that $r^2 = f^2+g^2-\frac{36}{5}$. Substituting the derived values for $f$ and $g$ allows us to solve for $\alpha$, resulting in $\alpha = \frac{36}{25} + \frac{169}{25} - \frac{180}{25} = 1$. Given that $r = \sqrt{\alpha} = 1$, the final sum $\alpha + r$ equals $2$.

The circle defined by $2x^2+2y^2-(1+a)x-(1-a)y=0$ can be written in standard form as $x^2+y^2-\frac{1+a}{2}x-\frac{1-a}{2}y=0$. The center of this circle is $C \left(\frac{1+a}{4}, \frac{1-a}{4}\right)$, and the point $P$ is given by $P \left(\frac{1+a}{2}, \frac{1-a}{2}\right)$, which simplifies to double the center coordinates, denoted as $P(2h, 2k)$ where $(h,k)$ is the center. Since the line $x+y=0$ bisects the chords, any chord's midpoint $M$ must lie on this line, taking the form $M(\lambda, -\lambda)$.

The equation of a chord with a given midpoint $(x_1, y_1)$ for a circle $S=0$ is $T=S_1$. For our circle $x^2+y^2-2hx-2ky=0$ and midpoint $(\lambda, -\lambda)$, the chord equation becomes $x\lambda - y\lambda - h(x+\lambda) - k(y-\lambda) = \lambda^2 + \lambda^2 - 2h\lambda - 2k(-\lambda)$. Simplifying this results in $x(\lambda-h) - y(\lambda+k) = 2\lambda^2 - h\lambda + k\lambda$. Substituting the coordinates of $P(2h, 2k)$ for $(x, y)$ gives the quadratic equation $2\lambda^2 + \lambda(3k - 3h) + 2(h^2 + k^2) = 0$.

For two distinct chords to pass through $P$, this quadratic in $\lambda$ must have two distinct real roots, meaning the discriminant $D = (3(k - h))^2 - 4(2)(2(h^2 + k^2))$ must be strictly greater than zero. This simplifies to the inequality $9(k-h)^2 - 16(h^2+k^2) > 0$, or $7k^2 + 18kh + 7h^2 < 0$. Substituting the values $h+k = \frac{1}{2}$, $h-k = \frac{a}{2}$, $h^2+k^2 = \frac{1+a^2}{8}$, and $hk = \frac{1-a^2}{16}$ into this expression yields $7\left(\frac{1+a^2}{8}\right) + 18\left(\frac{1-a^2}{16}\right) < 0$. Solving this final inequality leads to $\frac{14+14a^2+18-18a^2}{16} < 0$, which reduces to $32 - 4a^2 < 0$, identifying the set of all values as $a^2 \in (8, \infty )$.

For any chord subtending an angle $\theta$ at the center of a circle with radius $R$, the locus of its midpoints forms a smaller concentric circle with radius $r = R \cos(\frac{\theta}{2})$. Given the circle radius $R=2$, the relation $r = 2 \cos(\frac{\theta}{2})$ defines the radius of the locus for any such chord. Using the provided equation $r_1^2 = r_2^2 + r_3^2$, we calculate the values for the given angles: for $\theta_1 = \frac{\pi}{3}$, $r_1 = 2 \cos(\frac{\pi}{6}) = \sqrt{3}$, and for $\theta_3 = \frac{2\pi}{3}$, $r_3 = 2 \cos(\frac{\pi}{3}) = 1$. Substituting $r_1^2 = 3$ and $r_3^2 = 1$ into the relation yields $3 = r_2^2 + 1$, which simplifies to $r_2 = \sqrt{2}$. Solving $2 \cos(\frac{\theta_2}{2}) = \sqrt{2}$ leads to $\cos(\frac{\theta_2}{2}) = \frac{1}{\sqrt{2}}$, confirming that $\frac{\theta_2}{2} = \frac{\pi}{4}$ and $\theta_2 = \frac{\pi}{2}$.

Since $PR$ serves as the diameter of circle $C$ and point $Q$ lies on the circumference, the angle $\angle PQR$ is a right angle, which implies that the product of the slopes of lines $PQ$ and $QR$ is $-1$. Given the coordinates $P(-3, 2)$ and $Q(9, 10)$, the slope of $PQ$ is $\frac{10-2}{9-(-3)} = \frac{8}{12} = \frac{2}{3}$. Therefore, the slope of $QR$ must be the negative reciprocal, $-\frac{3}{2}$. Setting the slope of $QR$ equal to this value using $R(\alpha, 4)$ gives $\frac{4-10}{\alpha-9} = -\frac{3}{2}$, which simplifies to $\frac{-6}{\alpha-9} = -\frac{3}{2}$, leading to $\alpha - 9 = 4$ and thus $\alpha = 13$.

With $R(13, 4)$ identified, the center $O$ of the circle is the midpoint of the diameter $PR$, calculated as $O = (\frac{-3+13}{2}, \frac{2+4}{2}) = (5, 3)$. A tangent to a circle is perpendicular to the radius at the point of tangency. The slope of the radius $OQ$ is $\frac{10-3}{9-5} = \frac{7}{4}$, so the tangent at $Q$ has a slope of $-\frac{4}{7}$, resulting in the equation $y - 10 = -\frac{4}{7}(x-9)$, which simplifies to $4x + 7y = 106$. The slope of the radius $OR$ is $\frac{4-3}{13-5} = \frac{1}{8}$, making the tangent at $R$ have a slope of $-8$, which results in the equation $y - 4 = -8(x-13)$ or $8x + y = 108$.

To find the intersection point $S$ of these two tangents, substitute $y = 108 - 8x$ into the first equation: $4x + 7(108 - 8x) = 106$. This simplifies to $4x + 756 - 56x = 106$, or $-52x = -650$, resulting in $x = 12.5$. Substituting this $x$-value back gives $y = 108 - 8(12.5) = 108 - 100 = 8$, so the intersection point $S$ is $(12.5, 8)$. Finally, since $S$ lies on the line $2x - ky = 1$, we have $2(12.5) - 8k = 1$, which becomes $25 - 8k = 1$, yielding $8k = 24$ and thus $k = 3$.

The radius of a circle centered at $C$ that is tangent to the chord $AB$ is equal to the perpendicular distance from $C$ to the line $AB$. We begin by determining the equations of the tangents at points $A(4, -11)$ and $B(8, -5)$ on the circle defined by $x^2+y^2-3x+10y-15=0$. By using the standard tangent replacement rules-where $x^2$ becomes $xx_1$, $y^2$ becomes $yy_1$, $x$ becomes $\frac{x+x_1}{2}$, and $y$ becomes $\frac{y+y_1}{2}$-the tangent at $A$ is $4x - 11y - \frac{3}{2}(x+4) + 5(y-11) - 15 = 0$. Expanding this, we get $8x - 22y - 3x - 12 + 10y - 110 - 30 = 0$, which simplifies to the line $5x - 12y - 152 = 0$.

Applying the same procedure for the point $B(8, -5)$, the tangent equation is $8x - 5y - \frac{3}{2}(x+8) + 5(y-5) - 15 = 0$. This simplifies to $16x - 10y - 3x - 24 + 10y - 50 - 30 = 0$, resulting in $13x - 104 = 0$, or simply $x=8$. To find the coordinates of $C$, we substitute $x=8$ into the tangent equation at $A$, giving $5(8) - 12y - 152 = 0$, or $40 - 152 = 12y$, which leads to $-112 = 12y$, so $y = -28/3$. Thus, the intersection point $C$ is $(8, -28/3)$.

Next, we define the line $AB$ passing through $(4, -11)$ and $(8, -5)$. The slope of this line is calculated as $\frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$. Using the point-slope form $y - (-11) = \frac{3}{2}(x - 4)$, we obtain $2y + 22 = 3x - 12$, which rearranges to $3x - 2y - 34 = 0$. Finally, the radius $r$ is the perpendicular distance from $C(8, -28/3)$ to $3x - 2y - 34 = 0$, given by the formula $r = \frac{|3(8) - 2(-28/3) - 34|}{\sqrt{3^2 + (-2)^2}}$. The numerator simplifies to $|24 + 56/3 - 34| = |56/3 - 10| = 26/3$, and the denominator is $\sqrt{9+4} = \sqrt{13}$. Consequently, the radius is $r = \frac{26/3}{\sqrt{13}} = \frac{26}{3\sqrt{13}} = \frac{2\sqrt{13}}{3}$.

A circle located in the first quadrant and touching both coordinate axes has a center at $(r, r)$ and a radius $r$, meaning its equation is $(x-r)^2 + (y-r)^2 = r^2$. When such a circle intersects a line, the chord length is determined by $2\sqrt{r^2 - d^2}$, where $d$ is the perpendicular distance from the center $(r, r)$ to the line $x+y-2=0$.

Using the distance formula, $d = \frac{|r+r-2|}{\sqrt{1^2+1^2}}$, which simplifies to $\frac{|2r-2|}{\sqrt{2}}$. Equating the intercept length to 2 gives $2\sqrt{r^2 - d^2} = 2$, so $r^2 - d^2 = 1$. Substituting $d^2 = \frac{(2r-2)^2}{2} = 2(r-1)^2$ into the equation yields $r^2 - 2(r-1)^2 = 1$. This expands to $r^2 - 2(r^2 - 2r + 1) = 1$, which simplifies to the quadratic equation $r^2 - 4r + 3 = 0$.

The roots $r_1$ and $r_2$ represent the radii of the two circles. According to Vieta's formulas, $r_1+r_2 = 4$ and $r_1 r_2 = 3$. Evaluating the target expression $r_1^2 + r_2^2 - r_1 r_2$, which is $(r_1+r_2)^2 - 3r_1 r_2$, gives $4^2 - 3(3) = 16 - 9 = 7$.

The area of the triangle formed by the intersection point of two tangents to a circle depends on the radius of the circle and the length of the tangents drawn from that point. The circle given by $x^2 + y^2 - 2x + y - 5 = 0$ can be rewritten in the standard form $(x-1)^2 + (y + \frac{1}{2})^2 = \frac{25}{4}$, identifying the center as $(1, -\frac{1}{2})$ and the radius as $r = \frac{5}{2}$.

Evaluating the power of the point $R\left(\frac{9}{4}, 2\right)$ relative to the circle gives the square of the tangent length $L^2 = \left(\frac{9}{4}\right)^2 + 2^2 - 2\left(\frac{9}{4}\right) + 2 - 5 = \frac{25}{16}$, which means the tangent length is $L = \frac{5}{4}$.

The area of triangle $PQR$ is expressed by the formula
$\text{Area} = \frac{r L^3}{r^2 + L^2}$
Substituting $r = \frac{5}{2}$ and $L = \frac{5}{4}$ into this expression results in
$\text{Area} = \frac{(\frac{5}{2}) (\frac{5}{4})^3}{(\frac{5}{2})^2 + (\frac{5}{4})^2} = \frac{\frac{625}{128}}{\frac{25}{4} + \frac{25}{16}} = \frac{\frac{625}{128}}{\frac{125}{16}} = \frac{5}{8}$

The intersection points $A(\alpha, 0)$ and $B(1, \beta)$ of the line $ax+by=0$ and the circle $x^2+y^2-2x=0$ are identified as $A(0,0)$ and $B(1,1)$, as this is the only pair that satisfies the condition $a \neq b$. Using these endpoints to represent the diameter, the equation of the circle is determined by $(x-0)(x-1) + (y-0)(y-1) = 0$, which simplifies to $x^2+y^2-x-y=0$. This circle possesses a center at $(1/2, 1/2)$ and a radius squared of $1/2$. Since reflection across a line preserves the radius of a circle, the transformation requires only that we reflect the center across the line $x+y+2=0$. Applying the reflection formula $\frac{h'-1/2}{1} = \frac{k'-1/2}{1} = -2\frac{1/2+1/2+2}{1^2+1^2}$, we find the new center $(h', k')$ to be $(-5/2, -5/2)$. The equation of the resulting reflected circle is therefore $(x+5/2)^2 + (y+5/2)^2 = 1/2$, which expands to $x^2+y^2+5x+5y+12=0$.

The circumcircle of $\triangle OPQ$, where $OP$ and $OQ$ are tangents from the origin $O(0,0)$ to the circle $x^2+y^2-6x+4y+8=0$, has the line segment connecting the origin and the circle's center $C(3, -2)$ as its diameter because the radii to the points of tangency are perpendicular to the tangents. Using the endpoints of this diameter, the equation of the circumcircle is $(x-0)(x-3) + (y-0)(y+2) = 0$, which simplifies to $x^2+y^2-3x+2y=0$.

Since the circumcircle passes through the point $(\alpha, 1/2)$, substituting these coordinates into the equation gives $\alpha^2 + (1/2)^2 - 3\alpha + 2(1/2) = 0$. Simplifying this yields $\alpha^2 - 3\alpha + 1/4 + 1 = 0$, which can be written as $\alpha^2 - 3\alpha + 5/4 = 0$. Multiplying this equation by 4 results in the quadratic form $4\alpha^2 - 12\alpha + 5 = 0$, which factors as $(2\alpha-1)(2\alpha-5) = 0$. Solving this gives the possible values of $\alpha$ as $\alpha = 1/2$ and $\alpha = 5/2$.

The center of a circle $(\alpha, \beta)$ must satisfy the equation of the normal $4 \alpha + 3 \beta = 1$ because the normal line always passes through the center of the circle. Furthermore, the distance from this center to each tangent line must equal the radius $r$. By equating the perpendicular distances from the center $(\alpha, \beta)$ to the lines $3 x + 4 y - 24 = 0$ and $3 x - 4 y - 32 = 0$, the relationship $\left| \frac{3 \alpha + 4 \beta - 24}{5} \right| = \left| \frac{3 \alpha - 4 \beta - 32}{5} \right|$ is established.

Solving this equality for the case where the expressions share the same sign yields $3 \alpha + 4 \beta - 24 = 3 \alpha - 4 \beta - 32$, which simplifies to $8 \beta = -8$, identifying $\beta = -1$. Substituting this value into the normal equation $4 \alpha + 3 \beta = 1$ results in $4 \alpha - 3 = 1$, so $\alpha = 1$. The radius $r$ is determined by the distance from the center $(1, -1)$ to either tangent line, calculated as $\frac{|3(1) + 4(-1) - 24|}{5} = 5$. Since this result satisfies the condition $r < 8$, these coordinates are correct, as alternative solutions yield a radius exceeding the specified bound. The final value is computed as $\alpha - \beta + r = 1 - (-1) + 5 = 7$.