Circles – JEE Main MathematicsPractice Questions & PYQs

For a circle represented by $x^2+y^2-\alpha x+\beta y+\gamma=0$, the center is located at $(\frac{\alpha}{2}, -\frac{\beta}{2})$ and the radius $r$ satisfies $r^2 = (\frac{\alpha}{2})^2 + (\frac{\beta}{2})^2 - \gamma$. Tangency to the $x$-axis at $(a, 0)$ requires the center's $x$-coordinate to be $a$, implying $\alpha = 2a$, and the radius $r$ to equal the magnitude of the center's $y$-coordinate, which is $\frac{\beta}{2}$. Equating $r^2$ to the squared radius expression gives $\frac{\beta^2}{4} = \frac{\alpha^2}{4} + \frac{\beta^2}{4} - \gamma$, which simplifies to $\gamma = \frac{\alpha^2}{4}$. Setting $x=0$ to find the $y$-intercept of length $b$ yields the quadratic equation $y^2+\beta y+\gamma=0$, where the relationship between roots gives $b^2 = \beta^2 - 4\gamma$. Substituting these findings, the ordered pair $(2a, b^2)$ evaluates to $(\alpha, \beta^2 - 4\gamma)$.

The equation of the circle $C$ is $x^2 + (y-1)^2 = 2$, which has centre $(0, 1)$ and radius $\sqrt{2}$. The distance from the centre $(0, 1)$ to the line $x+y-3=0$ is $\frac{|0+1-3|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$, confirming that $x+y=3$ is tangent to $C$. The normal to the line at the point of contact $P(x_1, y_1)$ has slope $1$ and passes through $(0, 1)$, giving the equation $y=x+1$. Solving $y=x+1$ and $y=3-x$ yields the intersection point $P(1, 2)$, so $x_1 y_1 = 2$.

For the ellipse $E_1$ with equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the tangent $y = -x+3$ satisfies the condition $c^2 = a^2 m^2 + b^2$, where $c=3$ and $m=-1$, leading to $a^2+b^2=9$. The point of contact $Q(x_2, y_2)$ is $(\frac{a^2}{3}, \frac{b^2}{3})$. Similarly, for the ellipse $E_2$ with equation $\frac{x^2}{b_2^2} + \frac{y^2}{a_2^2} = 1$, the same condition $a_2^2+b_2^2=9$ holds, and the contact point $R(x_3, y_3)$ is $(\frac{b_2^2}{3}, \frac{a_2^2}{3})$. Since $P$ is the midpoint of $QR$, we have $x_2+x_3 = 2x_1 = 2$ and $y_2+y_3 = 2y_1 = 4$, which implies $\frac{a^2+b_2^2}{3} = 2$ and $\frac{b^2+a_2^2}{3} = 4$.

The distance $PQ^2$ is $(x_2-1)^2 + (y_2-2)^2 = (\frac{2\sqrt{2}}{3})^2 = \frac{8}{9}$. Substituting $x_2 = \frac{a^2}{3}$ and $y_2 = \frac{b^2}{3} = \frac{9-a^2}{3}$ yields $(\frac{a^2}{3}-1)^2 + (\frac{9-a^2}{3}-2)^2 = (\frac{a^2-3}{3})^2 + (\frac{3-a^2}{3})^2 = \frac{8}{9}$. This simplifies to $2(\frac{a^2-3}{3})^2 = \frac{8}{9}$, so $(a^2-3)^2 = 4$, which gives $a^2=5$ or $a^2=1$. Correspondingly, $b^2=4$ or $b^2=8$. Thus the coordinates for $Q$ and $R$ are $(\frac{5}{3}, \frac{4}{3})$ and $(\frac{1}{3}, \frac{8}{3})$ in some order. Calculating the product sum, $x_2 y_2 + x_3 y_3 = (\frac{5}{3})(\frac{4}{3}) + (\frac{1}{3})(\frac{8}{3}) = \frac{20}{9} + \frac{8}{9} = \frac{28}{9}$. Finally, $9(x_1 y_1 + x_2 y_2 + x_3 y_3) = 9(2 + \frac{28}{9}) = 18 + 28 = 46$.

The shortest distance from an external point to a circle is always found by calculating the straight-line distance from that point to the circle's center and then subtracting the radius. Because the circle has a radius of $1$ and touches the lines $x=3$ and $y=2$, its center must be $1$ unit away from each of these lines, placing it at $(3\pm 1, 2\pm 1)$. Comparing the potential centers of $(2,1), (4,1), (2,3),$ and $(4,3)$, the requirement that the circle be closer to the origin identifies $(2,1)$ as the correct center, leading to the equation $(x-2)^2 + (y-1)^2 = 1$.

Calculating the distance from the point $(5,5)$ to the center $(2,1)$ involves the standard distance formula, yielding $\sqrt{(5-2)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = 5$. Since this segment represents the distance to the center, we subtract the circle's radius of $1$ to determine the shortest gap between the point and the boundary of the circle, which is $5 - 1 = 4$.

To find the image of a point $(x_1, y_1)$ reflected across the line $ax+by+c=0$, we determine the new coordinates $(x', y')$ using the standard reflection property:

$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \left( \frac{ax_1 + by_1 + c}{a^2 + b^2} \right)$

Substituting the given point $(-4, 5)$ and the line $x+2y-2=0$ into this formula gives $\frac{x' + 4}{1} = \frac{y' - 5}{2} = -2 \left( \frac{-4 + 10 - 2}{1^2 + 2^2} \right)$. Evaluating the right side results in $-2 (\frac{4}{5})$, which is $-\frac{8}{5}$. Equating each part to this value, we solve for $x'$ and $y'$ to get $x' = -4 - \frac{8}{5} = -\frac{28}{5}$ and $y' = 5 - \frac{16}{5} = \frac{9}{5}$. Since this reflected point lies on the circle $(x+4)^2 + (y-3)^2 = r^2$, substituting these coordinates yields $(-\frac{28}{5} + 4)^2 + (\frac{9}{5} - 3)^2 = r^2$, which simplifies to $(-\frac{8}{5})^2 + (-\frac{6}{5})^2 = \frac{64}{25} + \frac{36}{25} = \frac{100}{25} = 4$. Thus, $r = 2$.

Two circles touching externally have a point of contact that divides the line segment joining their centers internally in the ratio of their respective radii. Given the point $(6,6)$ divides the segment between the centers $C_1(\alpha, \beta)$ and $C_2(8, \frac{15}{2})$ in a $2:1$ ratio, we can establish that the ratio of the radii $r_1:r_2$ is $2:1$, implying $r_1 = 2r_2$. Applying the section formula to these points gives the equations $\frac{2(8) + 1(\alpha)}{2+1} = 6$ and $\frac{2(\frac{15}{2}) + 1(\beta)}{2+1} = 6$, which solve to $\alpha=2$ and $\beta=3$.

The distance between the centers $C_1(2,3)$ and $C_2(8, \frac{15}{2})$ equals the sum of the radii $r_1 + r_2 = 3r_2$. Calculating this distance, we find $d = \sqrt{(8-2)^2 + (\frac{15}{2}-3)^2} = \sqrt{6^2 + (\frac{9}{2})^2} = \sqrt{36 + \frac{81}{4}} = \sqrt{\frac{144+81}{4}} = \frac{15}{2}$. Setting $3r_2 = \frac{15}{2}$ yields $r_2 = \frac{5}{2}$ and $r_1 = 5$. Substituting these values into the expression $(\alpha+\beta) + 4(r_1^2 + r_2^2)$, we get $(2+3) + 4(25 + \frac{25}{4}) = 5 + 4(\frac{125}{4}) = 5 + 125 = 130$.

The circle equation $x^2+y^2-16x-4y=0$ rearranges to $(x-8)^2+(y-2)^2=68$, identifying the center at $(8, 2)$. A line passing through this center with slope $m$ satisfies the equation $(y-2)=m(x-8)$. Setting $y=0$ yields the $x$-intercept $OA = 8-\frac{2}{m}$, while setting $x=0$ yields the $y$-intercept $OB = 2-8m$. Summing these segment lengths results in the expression $OA+OB = 10-\left(\frac{2}{m}+8m\right)$.

Because the line intersects the positive coordinate axes, the slope $m$ must be negative; substituting $m=-k$ for a positive constant $k$ transforms the sum into $10+\frac{2}{k}+8k$. Applying the AM-GM inequality to the variable terms ensures:
$\frac{2}{k}+8k \ge 2\sqrt{\frac{2}{k} \cdot 8k} = 2\sqrt{16} = 8$
The minimum total length is therefore $10+8=18$.

The inradius $r$ of an equilateral triangle with side length 12 is calculated as $r = \frac{12}{2\sqrt{3}} = 2\sqrt{3}$. Since a square inscribed in this circle must have a diagonal equal to the diameter $2r = 4\sqrt{3}$, one can relate the diagonal to the side length $s$ using the Pythagorean theorem, where $2s^2 = (4\sqrt{3})^2 = 48$. This calculation establishes the square's area as $m = s^2 = 24$. Given the side length $s = \sqrt{24} = 2\sqrt{6}$, the perimeter is $n = 4s = 8\sqrt{6}$. Calculating the sum $m + n^2 = 24 + (8\sqrt{6})^2$ leads to $24 + 384 = 408$.

Completing the square for the circle equation $x^2+y^2-10x-6y+30=0$ reveals the center at $(5, 3)$ and a radius of $r=2$. For a square inscribed in this circle, the perpendicular distance from the center to each side must be $r/\sqrt{2}$, which simplifies to $\sqrt{2}$. Given that the sides of the square are parallel to $y=x+3$ and perpendicular to each other, the lines containing these sides take the form $x-y+c=0$ and $x+y+d=0$.

Applying the perpendicular distance formula $|\frac{Ax_0+By_0+K}{\sqrt{A^2+B^2}}| = \sqrt{2}$ with the center $(5, 3)$ leads to the equations $|5-3+c|/\sqrt{2} = \sqrt{2}$ and $|5+3+d|/\sqrt{2} = \sqrt{2}$. Solving these absolute value equations determines that $c \in \{0, -4\}$ and $d \in \{-10, -6\}$. Intersecting these side lines identifies the four vertices of the square as $(5, 5)$, $(3, 3)$, $(7, 3)$, and $(5, 1)$. Summing the squared distances of these points from the origin, expressed as $(5^2+5^2) + (3^2+3^2) + (7^2+3^2) + (5^2+1^2)$, yields $50 + 18 + 58 + 26 = 152$.

The geometry of two circles touching relies on the relationship between their centers and radii. A circle passing through $(0,0)$ with center $(h,k)$ necessarily has a radius $r = \sqrt{h^2+k^2}$. Because the circle also passes through $(1,0)$, the center $(h,k)$ must be equidistant from $(0,0)$ and $(1,0)$, which places it on the perpendicular bisector $x=1/2$ and identifies $h = 1/2$. The circle $x^2+y^2=9$ is centered at $(0,0)$ with a radius of $3$. For the two circles to be tangent, the distance between their centers must equal the difference of their radii, expressed as $r = |3 - r|$. Solving this equation leads to $2r = 3$, which means $r = 3/2$. Since $r^2 = h^2+k^2$, it follows that $h^2+k^2 = (3/2)^2 = 9/4$. Consequently, the value of $4(h^2+k^2)$ equals $4 \times (9/4)$, resulting in $9$.

The orthocenter of a triangle is the intersection of its altitudes, meaning the segments from vertices $A(5, -2)$ and $B(8, 3)$ to the orthocenter $P(6, 1)$ are perpendicular to their opposite sides, with slopes calculated as $3$ and $1$, respectively. Consequently, sides $BC$ and $AC$ are perpendicular to these altitudes, resulting in slopes of $-\frac{1}{3}$ and $-1$. The equations for these sides, derived using these slopes and the known vertices, are $3y+x-17=0$ and $x+y-3=0$. Solving this system determines the coordinates of vertex $C$ to be $(-4, 7)$, which satisfies the circle equation $x^2+y^2-65=0$.

Finding the center of the circle involves identifying the point $(x_c, y_c)$ where the two lines $3x+5y=1$ and $(2+c)x+5c^2y=1$ intersect. By solving the first equation for $y$ as $y = \frac{1-3x}{5}$ and substituting this into the second equation, we obtain $(2+c)x + 5c^2(\frac{1-3x}{5}) = 1$, which simplifies to $(2+c)x + c^2 - 3c^2x = 1$. Rearranging this expression to isolate $x$ yields $x(2+c-3c^2) = 1-c^2$. Factoring the quadratic term $2+c-3c^2$ as $(3c+2)(1-c)$ and the numerator $1-c^2$ as $(1-c)(1+c)$, we can determine the coordinate $x_c = \frac{1+c}{3c+2}$.

Taking the limit as $c$ approaches $1$ allows us to find the coordinates of the circle's center $(h, k)$. For $h$, the limit $\lim_{c\to 1} \frac{1+c}{3c+2}$ results in $\frac{2}{5}$. Substituting this value into the first linear equation $3(\frac{2}{5}) + 5y = 1$ gives $5y = 1 - \frac{6}{5}$, which simplifies to $5y = -\frac{1}{5}$, so $y_c = -\frac{1}{25}$. Thus, the center of the circle is located at $(\frac{2}{5}, -\frac{1}{25})$.

The circle passes through the point $(2, 0)$, so the square of the radius $r^2$ is the squared distance between the center $(\frac{2}{5}, -\frac{1}{25})$ and the point $(2, 0)$. This gives $r^2 = (2 - \frac{2}{5})^2 + (0 - (-\frac{1}{25}))^2$, which is $(\frac{8}{5})^2 + (\frac{1}{25})^2$, equating to $\frac{64}{25} + \frac{1}{625} = \frac{1600+1}{625} = \frac{1601}{625}$. Using the standard circle equation $(x-h)^2 + (y-k)^2 = r^2$, we write $(x - \frac{2}{5})^2 + (y + \frac{1}{25})^2 = \frac{1601}{625}$. Expanding this yields $x^2 - \frac{4}{5}x + \frac{4}{25} + y^2 + \frac{2}{25}y + \frac{1}{625} = \frac{1601}{625}$. Multiplying the entire equation by $25$ results in $25x^2 - 20x + 4 + 25y^2 + 2y + \frac{1}{25} = \frac{1601}{25}$, and further simplification confirms the equation $25x^2+25y^2-20x+2y-60=0$.

Two circles intersect at exactly two distinct points when the distance between their centers is greater than the absolute difference of their radii but less than the sum of their radii. By converting the second circle equation to standard form $(x-2)^2 + (y-2)^2 = 4$, we identify its center as $(2, 2)$ and its radius as $2$, while the first circle has center $(-1, -2)$ and radius $r$. The distance $d$ between these centers is $\sqrt{(2 - (-1))^2 + (2 - (-2))^2} = 5$. Setting up the required inequality $|r - 2| < 5 < r + 2$ breaks down into the conditions $r > 3$ and $r < 7$, which together define the range $3 < r < 7$.

The midpoint $M(h,k)$ of any chord of the circle $x^2 + (y-1)^2 = 1$ that passes through the origin must satisfy the property that the segment connecting the circle's center $C(0,1)$ to $M$ is perpendicular to the chord itself. Equating the product of the slopes of the lines $OM$ and $CM$ to $-1$ gives $\frac{k}{h} \cdot \frac{k-1}{h} = -1$, which simplifies to the locus equation $x^2 + y^2 - y = 0$. To find the points $P$ and $Q$, we substitute the line equation $y = 1-x$ into the locus equation, resulting in $x^2 + (1-x)^2 - (1-x) = 0$, which reduces to $2x^2 - x = 0$. Solving this yields the intersection coordinates $(0,1)$ and $(1/2, 1/2)$, and calculating the distance between these points provides $PQ = \sqrt{(1/2-0)^2 + (1/2-1)^2} = \sqrt{1/4+1/4} = \frac{1}{\sqrt{2}}$.

For the circle $x^2+y^2-10 x+4 y+13=0$, we first determine its center and radius by completing the square to express the equation as $(x-5)^2+(y+2)^2=16$. This identifies the circle's center at $(5, -2)$ and its radius as $4$. Since a diameter of this circle acts as a chord for circle $C$, the length of that chord must equal the diameter of the first circle, which is $2 \times 4 = 8$.

We find the center of circle $C$ by solving the system of linear equations $2 x+3 y=12$ and $3 x-2 y=5$. Multiplying the first equation by $2$ and the second by $3$ allows for the elimination of $y$, yielding $13 x=39$, so $x=3$. Substituting this value into the equation $2 x+3 y=12$ gives $6+3 y=12$, which simplifies to $3 y=6$, or $y=2$. Thus, the center of circle $C$ is located at $(3, 2)$.

The distance $d$ between the center of the given circle $(5, -2)$ and the center of circle $C$ $(3, 2)$ acts as the perpendicular distance from the center of circle $C$ to the common chord. Using the distance formula, $d = \sqrt{(5-3)^2+(-2-2)^2}$, which results in $\sqrt{4+16}=\sqrt{20}$. Within circle $C$, the radius $R$ forms a right triangle with this distance $d$ and half the chord length $a=4$. Applying the Pythagorean theorem, $R^2=d^2+a^2$, which gives $R^2=20+16=36$. Consequently, the radius of circle $C$ is $6$.

Since any three non-collinear points uniquely define a circle, the equation of the circle passing through the points $(0,0)$, $(1,0)$, and $(0,1)$ can be determined by observing that they satisfy the equation $x(x-1) + y(y-1) = 0$. Expanding this expression results in the standard form $x^2 + y^2 - x - y = 0$, which defines the locus of the circle.

For all four points to lie on this circle, the coordinates $(2k, 3k)$ must also satisfy the derived equation. Substituting these values into the expression yields $(2k)^2 + (3k)^2 - 2k - 3k = 0$, which simplifies to $13k^2 - 5k = 0$. Factoring the resulting quadratic equation gives $k(13k - 5) = 0$, resulting in potential solutions of $k = 0$ and $k = \frac{5}{13}$. Because the problem requires four distinct points, $k = 0$ must be excluded as it makes the fourth point coincide with the origin, leaving $k = \frac{5}{13}$ as the only valid solution.

Two circles intersect at two distinct points if the distance between their centers, $d$, satisfies the strict inequality $|r_1 - r_2| < d < r_1 + r_2$. The circle $C: x^2 + y^2 = 4$ has a center at $(0, 0)$ and a radius $r_1 = 2$. Rewriting the second circle $C': x^2 + y^2 - 4\lambda x + 9 = 0$ as $(x - 2\lambda)^2 + y^2 = 4\lambda^2 - 9$ reveals a center at $(2\lambda, 0)$ and a radius $r_2 = \sqrt{4\lambda^2 - 9}$, which exists only when $4\lambda^2 - 9 > 0$, or $|\lambda| > \frac{3}{2}$.

Substituting these values into the intersection condition produces $|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$. Squaring the right part of this inequality, $2|\lambda| < 2 + \sqrt{4\lambda^2 - 9}$, simplifies to $4\lambda^2 < 4 + 4\lambda^2 - 9 + 4\sqrt{4\lambda^2 - 9}$, which reduces to $5 < 4\sqrt{4\lambda^2 - 9}$. Squaring both sides yields $25 < 16(4\lambda^2 - 9)$, leading to $25 < 64\lambda^2 - 144$, which simplifies to $169 < 64\lambda^2$, or $\lambda^2 > (\frac{13}{8})^2$. This condition defines the valid set of $\lambda$ values as $(-\infty , -\frac{13}{8}) \cup (\frac{13}{8}, \infty )$, equivalent to the set $R - [-\frac{13}{8}, \frac{13}{8}]$.

Identifying $a = -\frac{13}{8}$ and $b = \frac{13}{8}$ allows for the determination of the point $(8a + 12, 16b - 20)$. Calculating these coordinates gives $x = 8(-\frac{13}{8}) + 12 = -1$ and $y = 16(\frac{13}{8}) - 20 = 6$. Substituting $(-1, 6)$ into the equation $6x^2 + y^2 = 42$ results in $6(-1)^2 + (6)^2 = 6 + 36 = 42$, confirming that the point lies on this curve.

The equation of a common chord between two circles is found by calculating the difference between their respective equations, expressed as $S_1 - S_2 = 0$. The circle $C_1$ is defined by $x^2+y^2-2x-2y+1=0$, which implies a center at $(1,1)$. For $C_2$, with a center at $(-1,0)$ and radius 2, the equation is $(x+1)^2+y^2=4$, which expands to $x^2+y^2+2x-3=0$. Subtracting the second equation from the first, $(x^2+y^2-2x-2y+1) - (x^2+y^2+2x-3)=0$, results in the linear relationship $-4x-2y+4=0$, which simplifies to $2x+y=2$.

Point $P$ lies on the $y$-axis where $x=0$. Substituting this value into $2x+y=2$ results in $y=2$, identifying the coordinates of $P$ as $(0,2)$. The squared distance between $P(0,2)$ and the center of $C_1$ at $(1,1)$ is determined by $(1-0)^2 + (1-2)^2$, which calculates to $1^2 + (-1)^2 = 2$.