Circles – JEE Main MathematicsPractice Questions & PYQs

For any circle defined by the general equation $x^{2}+y^{2}+g x+f y+c=0$, the radius is given by the formula $r=\sqrt{g^{2}/4+f^{2}/4-c}$. Applying this to the provided equation $x^{2}+y^{2}+p x+(1-p) y+5=0$, the radius is $r=\sqrt{p^{2}/4+(1-p)^{2}/4-5}$, which simplifies to $r=\sqrt{(2p^{2}-2p-19)/4}$. Given that the radius must satisfy $0 < r \leq 5$, we square these bounds to obtain the inequality $0 < (2p^{2}-2p-19)/4 \leq 25$, or equivalently, $0 < 2p^{2}-2p-19 \leq 100$.

Solving the left side of this compound inequality, $2p^{2}-2p-19 > 0$, establishes the lower boundary for the value of $p^{2}$. Specifically, the critical values derived from this quadratic lead to the requirement that $p^{2}$ must be greater than approximately $6.88$. Conversely, the right side, $2p^{2}-2p-19 \leq 100$, simplifies to $2p^{2}-2p-119 \leq 0$, which sets an upper limit on $p^{2}$ at approximately $67.72$. Since $q=p^{2}$ is defined as an integer, these boundaries restrict $q$ to the set of integers from $7$ up to $67$ inclusive. Counting the number of integers in this range, from $7$ to $67$ inclusive, yields a total of $61$ elements.

The circle defined by the equation $x^{2}+y^{2}-2 x-6 y+6=0$ has a center located at $(1, 3)$ and a radius of $2$ units, which is determined by rewriting the equation in the standard form $(x-1)^{2}+(y-3)^{2}=4$. Because a diameter of this circle serves as a chord for a second circle centered at $(2, 1)$, these components establish a right-angled triangle where the radius of the second circle acts as the hypotenuse.

The distance between the center of the first circle $(1, 3)$ and the center of the second circle $(2, 1)$ is calculated as $\sqrt{(2-1)^{2}+(1-3)^{2}}$, which simplifies to $\sqrt{1+4}=\sqrt{5}$ units. According to the Pythagorean theorem, the square of the radius of the second circle is equal to the sum of the square of the first circle's radius and the square of the distance between the centers. Substituting these values yields $2^{2}+(\sqrt{5})^{2}=4+5=9$, meaning the radius of the second circle is $3$.

To analyze the geometry of these circles, we first rewrite their equations in the standard form $(x-h)^2 + (y-k)^2 = r^2$. For the first circle, completing the square for the $x$ and $y$ terms yields $(x-5)^2 + (y-5)^2 = 9$, identifying a center at (5, 5) with a radius of 3. Applying the same process to the second equation, $x^2+y^2-16x-10y+80=0$, results in $(x-8)^2 + (y-5)^2 = 9$, which shows this circle also has a radius of 3 but a center at (8, 5).

The distance between these centers is calculated using the distance formula, resulting in $\sqrt{(8-5)^2 + (5-5)^2} = 3$. Since the average of the two radii, $\frac{3+3}{2} = 3$, also equals 3, the distance between the centers is exactly equal to the radius of the circles. Substituting the center of the first circle into the equation of the second, and vice-versa, confirms that both centers lie directly on the circumference of the other circle, rather than within their respective interior regions. The claim that the centers lie inside the region of one another is therefore incorrect, as the points lie on the boundaries.

To determine how these two circles interact, we must examine the relationship between the distance separating their centers and their respective radii. By converting the equations into standard form, we find the first circle is centered at $(5, 5)$ with a radius of $3$, while the second circle is centered at $(11, 5)$ with a radius of $3$. The distance between these two centers is calculated as $\sqrt{(11-5)^{2}+(5-5)^{2}}$, which results in a value of $6$. Since this distance is exactly equal to the sum of the radii, $3 + 3 = 6$, the circles touch each other externally at exactly one point.

The equation of the tangent line to the circle $x^{2}+y^{2}=25$ at the point $R(3,4)$ is determined by the formula $xx_{1}+yy_{1}=25$, which yields $3x+4y=25$. Evaluating this equation at $y=0$ identifies the x-intercept $P$ at $(\frac{25}{3}, 0)$, while setting $x=0$ provides the y-intercept $Q$ at $(0, \frac{25}{4})$. These intercepts define a right-angled triangle $OPQ$ with leg lengths of $\frac{25}{3}$ and $\frac{25}{4}$.

The hypotenuse $PQ$ of this triangle is calculated using the Pythagorean theorem as $\sqrt{(\frac{25}{3})^{2}+(\frac{25}{4})^{2}} = \frac{125}{12}$. For a right-angled triangle with legs $a$ and $b$ and hypotenuse $c$, the incentre is located at $(r, r)$, where the inradius $r$ is given by $\frac{a+b-c}{2}$. Substituting the geometric dimensions gives $r = \frac{\frac{25}{3} + \frac{25}{4} - \frac{125}{12}}{2}$, which simplifies to $\frac{25}{12}$. Since the required circle is centered at the incentre $(\frac{25}{12}, \frac{25}{12})$ and passes through the origin $(0,0)$, its squared radius $r^{2}$ is the square of the distance from the origin to the center, calculated as $(\frac{25}{12})^{2} + (\frac{25}{12})^{2} = 2 \times \frac{625}{144} = \frac{625}{72}$.

The family of circles tangent to the line $4x+3y=10$ at the point $(1,2)$ is given by the equation $(x-1)^2+(y-2)^2+\lambda(4x+3y-10)=0$. Expanding and rearranging this into the standard form $x^2+y^2+2gx+2fy+c=0$ identifies the center as $(-(2\lambda-1), -(1.5\lambda-2))$ and the constant term as $c=5-10\lambda$. Since the radius of the circles is $5$, the condition $r^2=g^2+f^2-c$ becomes $(2\lambda-1)^2+(1.5\lambda-2)^2-(5-10\lambda)=25$. Solving this equation results in $6.25\lambda^2=25$, which simplifies to $\lambda^2=4$ or $\lambda=\pm 2$. Using $\lambda=2$, the center $C_1(\alpha, \beta)$ is determined to be $(-3,-1)$, and using $\lambda=-2$, the center $C_2(\gamma, \delta)$ is found to be $(5,5)$. The final value $|(\alpha+\beta)(\gamma+\delta)|$ is calculated as $|(-3-1)(5+5)|$, which equals $40$.

The shortest distance between two circles is obtained by subtracting the sum of their radii from the distance between their centers. Begin by expressing each equation in the standard form $(x-h)^{2}+(y-k)^{2}=r^{2}$ to isolate the center coordinates and radius. The first circle, $x^{2}+y^{2}-10x-10y+41=0$, rearranges to $(x-5)^{2}+(y-5)^{2}=3^{2}$, revealing a center at (5, 5) with a radius of 3. The second circle, $x^{2}+y^{2}-24x-10y+160=0$, becomes $(x-12)^{2}+(y-5)^{2}=3^{2}$, placing its center at (12, 5) with a radius of 3. The distance between the centers is calculated as $\sqrt{(12-5)^{2}+(5-5)^{2}}=7$. Subtracting the total of the two radii, which is 6, from this center-to-center distance of 7 yields a minimum distance of 1.

A circle touching the $y$-axis at $(0,6)$ implies its center has a $y$-coordinate of 6 and a horizontal distance from the $y$-axis equal to the radius $r$, placing the center at $(r, 6)$. The perpendicular distance from this center to the $x$-axis is 6, forming one leg of a right-angled triangle where the hypotenuse is the radius $r$. Since the circle cuts an intercept of $6\sqrt{5}$ on the $x$-axis, the line segment from the center to the chord's midpoint bisects this length, creating a base of $3\sqrt{5}$ for that same triangle. Applying the Pythagorean theorem to this configuration leads to the equation $r^2 = 6^2 + (3\sqrt{5})^2$, which simplifies to $r^2 = 36 + 45 = 81$, confirming that the radius is 9.

For one disk to fully contain another, the distance between their centers added to the radius of the smaller disk must be less than or equal to the radius of the larger disk. We begin by identifying the center and radius for each of the given sets by completing the square for the equations defining $A$, $B$, and $C$. For set $A$, the equation $2x^2+2y^2-2x-2y=1$ simplifies to $x^2+y^2-x-y=1/2$, which rearranges to $(x-1/2)^2 + (y-1/2)^2 = 1$, representing a disk with center $C_1(1/2, 1/2)$ and radius $R_1=1$. For set $B$, the equation $4x^2+4y^2-16y+7=0$ simplifies to $x^2+y^2-4y+7/4=0$, resulting in $(x-0)^2 + (y-2)^2 = 9/4$, which gives center $C_2(0, 2)$ and radius $R_2=3/2$. The set $C$ is defined by $(x-2)^2 + (y-1)^2 \le r^2$, a disk centered at $C_3(2, 1)$ with radius $R_3=|r|$.

Applying the geometric condition for containment, set $C$ must encompass both $A$ and $B$, which requires $d(C_1, C_3) + R_1 \le R_3$ and $d(C_2, C_3) + R_2 \le R_3$. We calculate the distances between the center of $C$ and the centers of $A$ and $B$ as $d(C_1, C_3) = \sqrt{(2-1/2)^2 + (1-1/2)^2} = \sqrt{(3/2)^2 + (1/2)^2} = \sqrt{9/4+1/4} = \sqrt{10/4} = \sqrt{5/2}$ and $d(C_2, C_3) = \sqrt{(2-0)^2 + (1-2)^2} = \sqrt{2^2+(-1)^2} = \sqrt{5}$. Substituting these values into our containment inequalities, we obtain $r \ge \sqrt{5/2} + 1$ and $r \ge \sqrt{5} + 3/2$. To satisfy both conditions simultaneously, $r$ must be greater than or equal to the maximum of these two values. Since $\sqrt{5} + 3/2$ is larger than $\sqrt{5/2} + 1$, the minimum radius is $\frac{3}{2} + \sqrt{5}$, which simplifies to $\frac{3+2\sqrt{5}}{2}$.

The locus of a point whose sum of distances to two fixed points remains constant is defined as an ellipse. Here, the fixed circles $S_{1}$ centered at $(0,0)$ with radius $3$ and $S_{2}$ centered at $(2,0)$ with radius $1$ act as the focal references. When a variable circle with center $P(h, k)$ and radius $r$ touches $S_{1}$ internally, the distance between centers is $PC_{1} = 3 - r$, and touching $S_{2}$ externally implies $PC_{2} = 1 + r$. Summing these expressions, we obtain $PC_{1} + PC_{2} = 4$, which provides the major axis length of $2a = 4$, meaning $a = 2$.

Because the distance between the foci $C_{1}$ and $C_{2}$ is $2$, the relationship $2ae = 2$ gives an eccentricity of $e = \frac{1}{2}$. The center of this ellipse is the midpoint of the foci, located at $(1,0)$, and the semi-minor axis squared follows from $b^{2} = a^{2}(1 - e^{2})$, resulting in $b^{2} = 3$. This leads to the equation of the locus:
$\frac{(x-1)^{2}}{4} + \frac{y^{2}}{3} = 1$
Substituting the coordinates $\left(2, \pm \frac{3}{2}\right)$ into this equation yields $\frac{1}{4} + \frac{3}{4} = 1$, confirming that these points satisfy the condition.

Converting each circle equation to the standard form $(x-h)^2 + (y-k)^2 = r^2$ reveals the coordinates of the centers. Circle $M$, defined by $x^2+y^2=1$, has a center at $(0,0)$. Circle $N$, given by $x^2+y^2-2x=0$, rearranges to $(x-1)^2+y^2=1$, identifying its center at $(1,0)$. Circle $O$, expressed as $x^2+y^2-2x-2y+1=0$, simplifies to $(x-1)^2+(y-1)^2=1$, placing its center at $(1,1)$. Circle $P$, represented by $x^2+y^2-2y=0$, converts to $x^2+(y-1)^2=1$, which places its center at $(0,1)$.

These four points-$(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$-form a quadrilateral where every side measures exactly $1$ unit in length. Because the lengths of the diagonals, connecting the opposite vertices $(0,0)$ to $(1,1)$ and $(1,0)$ to $(0,1)$, are both equal to $\sqrt{2}$, the figure defined by these centers is a square.

The circle defined by $x^{2}+y^{2}+2 x+4 y-4=0$ is equivalent to $(x+1)^2+(y+2)^2=3^2$, which places its center at $C(-1, -2)$ with a radius of $R=3$. The distance between this center and the fixed point $P(-4,1)$ is calculated as $CP = \sqrt{(-4 - (-1))^2 + (1 - (-2))^2} = 3\sqrt{2}$. Any circle passing through $P$ with its center on the circumference of the fixed circle will have a radius equal to the distance from that center to $P$, so the largest radius is $r_1 = CP + R = 3\sqrt{2} + 3$ and the smallest is $r_2 = CP - R = 3\sqrt{2} - 3$. The ratio of these radii simplifies as $\frac{r_1}{r_2} = \frac{3\sqrt{2}+3}{3\sqrt{2}-3} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$, and after rationalizing the denominator, this becomes $(\sqrt{2}+1)^2 = 3 + 2\sqrt{2}$. Since this is in the form $a+b\sqrt{2}$, we identify $a=3$ and $b=2$, resulting in a sum $a+b=5$.

The problem involves a circle centered at $C(2, 3)$ passing through the origin $O(0, 0)$. The radius of this circle is the distance between these two points, which is $r = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{13}$. Because the line segments $CP$ and $CQ$ are perpendicular to $OC$, the slope of the line passing through $C$, $P$, and $Q$ must be the negative reciprocal of the slope of $OC$. The slope of $OC$ is $\frac{3-0}{2-0} = \frac{3}{2}$, meaning the slope of the line segment containing $P$ and $Q$ is $-\frac{2}{3}$.

To find the coordinates of these points, we use the parametric form of a line extending from the center $C(2, 3)$ at a slope $\tan \theta = -\frac{2}{3}$. This gives us the coordinates $x = 2 \pm r \cos \theta$ and $y = 3 \pm r \sin \theta$. With a slope of $-\frac{2}{3}$, we determine that $\cos \theta = \pm \frac{3}{\sqrt{13}}$ and $\sin \theta = \mp \frac{2}{\sqrt{13}}$. Substituting $r = \sqrt{13}$ into the parametric equations yields $x = 2 \pm \sqrt{13} \left(\frac{3}{\sqrt{13}}\right)$ and $y = 3 \mp \sqrt{13} \left(\frac{2}{\sqrt{13}}\right)$. Simplifying these calculations leads to $x = 2 \pm 3$ and $y = 3 \mp 2$, which identifies the two points as $(-1, 5)$ and $(5, 1)$.

The locus of a point is governed by the standard distance formula, which states that the square of the distance between a moving point $(x, y)$ and any fixed point $(x_0, y_0)$ is given by $(x-x_0)^2 + (y-y_0)^2$. Applying this to the four given points $(0,0), (1,0), (0,1),$ and $(1,1)$, we set the sum of these squared distances equal to $18$. This results in the following equation:

$(x^2 + y^2) + ((x-1)^2 + y^2) + (x^2 + (y-1)^2) + ((x-1)^2 + (y-1)^2) = 18$

Expanding the squared terms and combining like variables simplifies the relationship to $4x^2 + 4y^2 - 4x - 4y + 4 = 18$. Dividing the entire equation by $4$ gives $x^2 + y^2 - x - y = 3.5$. To find the geometric properties, we complete the square for both $x$ and $y$, allowing us to rewrite the equation in the standard form $(x - 0.5)^2 + (y - 0.5)^2 = 4$. This confirms the radius of the circle is $2$, meaning the diameter $d$ is $4$, which results in $d^2 = 16$.

The geometry of a circle dictates that the normal at point $P(5,7)$ passes through the center $(2,3)$, while the tangent is perpendicular to this normal at $P$. The slope of the normal is $\frac{7-3}{5-2} = \frac{4}{3}$, which defines the line equation $4x-3y+1=0$. Because the tangent is perpendicular to the normal, its slope is $-\frac{3}{4}$, and applying the point-slope form results in the equation $3x+4y=43$.

These lines intersect the x-axis to form the base of the triangle. By setting $y=0$ in the normal equation, the x-intercept is found to be $-\frac{1}{4}$, and setting $y=0$ in the tangent equation gives an x-intercept of $\frac{43}{3}$. The base length along the x-axis is the distance between these points, calculated as $\frac{43}{3} - (-\frac{1}{4}) = \frac{175}{12}$. With the triangle's height equal to the y-coordinate of $P$, which is $7$, the area $A$ is determined by $\frac{1}{2} \times \frac{175}{12} \times 7$. Calculating $24A$ involves multiplying this result by 24, which simplifies to $24 \times \frac{1}{2} \times \frac{175}{12} \times 7 = 1225$.

The locus of a point moving such that the ratio of its distances from two fixed points is constant follows the geometric principle behind the Circle of Apollonius. By denoting the moving point as $P(x, y)$, we express its squared distance from $(5, 0)$ as $(x-5)^2 + y^2$ and its squared distance from $(-5, 0)$ as $(x+5)^2 + y^2$. Given the condition that the first distance is triple the second, we equate these squared distances as $(x-5)^{2}+y^{2}=9((x+5)^{2}+y^{2})$

Expanding both sides of this equation yields $x^2 - 10x + 25 + y^2 = 9x^2 + 90x + 225 + 9y^2$. Rearranging all terms to one side provides $8x^2 + 8y^2 + 100x + 200 = 0$, which simplifies to $x^2 + y^2 + \frac{25}{2}x + 25 = 0$ after dividing by eight. Comparing this to the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = \frac{25}{4}$ and $c = 25$.

Since the radius squared is defined by the formula $r^2 = g^2 - c$, we calculate $r^2 = \left(\frac{25}{4}\right)^2 - 25 = \frac{625}{16} - 25 = \frac{225}{16}$. Multiplying this result by four to obtain the required value gives $4r^2 = 4 \times \frac{225}{16} = 56.25$.

Lines tangent to the parabola $y^{2}=30x$ follow the general equation $y=mx+\frac{30}{4m}$. Given the chord passes through the point $(-30,0)$, substituting these coordinates results in $0=-30m+\frac{30}{4m}$, which simplifies to $m^{2}=\frac{1}{4}$ and yields slopes of $m=\pm\frac{1}{2}$. These tangents are expressed by the lines $x-2y+30=0$ and $x+2y+30=0$.

Dividing the circle equation $4x^{2}+4y^{2}+120x+675=0$ by four results in $x^{2}+y^{2}+30x+\frac{675}{4}=0$, identifying a center at $(-15,0)$ and a squared radius of $r^{2}=15^{2}-\frac{675}{4}=\frac{225}{4}$. The perpendicular distance $p$ from the center $(-15,0)$ to the tangent $x-2y+30=0$ is calculated as $p=\frac{|-15+30|}{\sqrt{1^{2}+(-2)^{2}}}=\frac{15}{\sqrt{5}}=3\sqrt{5}$. The chord length is determined by $2\sqrt{r^{2}-p^{2}}$, and substituting the values $r^{2}=\frac{225}{4}$ and $p^{2}=45$ yields
$2\sqrt{\frac{225}{4}-45}=2\sqrt{\frac{45}{4}}=3\sqrt{5}$

Any point on the circle defined by $(x-1)^2 + (y-1)^2 = 1$ can be represented using the parametric coordinates $P(1+\cos\theta, 1+\sin\theta)$. By calculating the sum of the squared distances to $A(1, 4)$ and $B(1, -5)$, we obtain $(PA)^2 + (PB)^2 = \cos^2\theta + (\sin\theta - 3)^2 + \cos^2\theta + (\sin\theta + 6)^2$, which simplifies to $2(\cos^2\theta + \sin^2\theta) + 6\sin\theta + 45$, or $47 + 6\sin\theta$. This sum reaches its maximum value when $\sin\theta = 1$, identifying the point $P$ as $(1, 2)$. Since the $x$-coordinates of $P(1, 2)$, $A(1, 4)$, and $B(1, -5)$ are identical, these three points all lie on the vertical line $x=1$ and are consequently collinear.

The locus of the midpoint of a line segment connecting a fixed point to a point on a circle forms another circle, scaled by a factor of one-half. Given the circle $x^{2} + y^{2} = 1$, any point $(a, b)$ on its boundary satisfies $a^{2} + b^{2} = 1$, while the midpoint $(h, k)$ of the segment from $(3, 2)$ is defined by the relations $a = 2h - 3$ and $b = 2k - 2$.

Substituting these coordinates into the original circle equation results in $(2h - 3)^{2} + (2k - 2)^{2} = 1$. Dividing both sides by $4$ produces the standard equation for the new circle:
$(h - \frac{3}{2})^{2} + (k - 1)^{2} = \frac{1}{4}$
Since the standard form is $(x - h_{0})^{2} + (y - k_{0})^{2} = r^{2}$, it follows that $r^{2} = \frac{1}{4}$, which confirms that the radius $r$ is $\frac{1}{2}$.

The principle behind this problem involves the family of circles passing through a specific point of tangency on a given line. Since circle $C$ touches the line $x-2y=0$ at the point $(2,1)$, its equation can be represented as $(x-2)^{2}+(y-1)^{2}+\lambda (x-2y)=0$. Expanding this expression leads to the general form $x^2 + y^2 + (\lambda-4)x + (-2-2\lambda)y + 5 = 0$.

When circle $C$ intersects the given circle $C_{1}: x^{2}+y^{2}+2y-5=0$, the common chord $PQ$ is found by subtracting the two circle equations. This subtraction results in the equation $(\lambda-4)x + (-4-2\lambda)y + 10 = 0$. Because the chord $PQ$ is a diameter of $C_{1}$, it must pass through the center of $C_{1}$. The center of $C_{1}$ is determined by the coefficients of $x$ and $y$ in its equation, which places it at $(0,-1)$. Substituting these center coordinates into the equation of the chord gives $0(\lambda-4) - 1(-4-2\lambda) + 10 = 0$, which simplifies to $4 + 2\lambda + 10 = 0$, resulting in $\lambda = -7$.

With the value of $\lambda$ determined, the full equation of circle $C$ becomes $x^2 + y^2 - 11x + 12y + 5 = 0$. For a circle in the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, the radius is calculated using $\sqrt{g^2 + f^2 - c}$. Here, $g = -5.5$ and $f = 6$, so the radius is $\sqrt{(-5.5)^2 + 6^2 - 5}$, which simplifies to $\sqrt{30.25 + 36 - 5}$ or $\sqrt{61.25}$. Expressed as a fraction, this is $\sqrt{\frac{245}{4}} = \frac{\sqrt{245}}{2}$. The diameter is twice the radius, giving $\sqrt{245}$, which simplifies to $7 \sqrt{5}$.