Circles – JEE Main MathematicsPractice Questions & PYQs

To determine the range of values for $\alpha$, we first identify the properties of the two circles given by the equations $(x-1)^{2}+(y-1)^{2}=1$ and $(x-9)^{2}+(y-1)^{2}=4$. The first circle has a center at $(1, 1)$ with a radius $r_1=1$, and the second circle has a center at $(9, 1)$ with a radius $r_2=2$. For a variable line $3x+4y-\alpha=0$ to lie between these two circles without intercepting a chord, it must satisfy two conditions: the centers must lie on opposite sides of the line, and the perpendicular distance from each center to the line must be at least the radius of the respective circle.

The condition for the centers $(1, 1)$ and $(9, 1)$ to lie on opposite sides of the line is that the product of the values obtained by substituting the center coordinates into the line equation must be negative, expressed as $(3(1)+4(1)-\alpha)(3(9)+4(1)-\alpha) < 0$. This simplifies to $(7-\alpha)(31-\alpha) < 0$, which defines the interval $\alpha \in (7, 31)$.

Next, we calculate the perpendicular distance from each center to the line $3x+4y-\alpha=0$. The distance from $(1, 1)$ is given by $\frac{|3(1)+4(1)-\alpha|}{\sqrt{3^2+4^2}} = \frac{|7-\alpha|}{5}$. For the line not to intersect the first circle, we require $\frac{|7-\alpha|}{5} \geq 1$, which implies $|7-\alpha| \geq 5$, resulting in $\alpha \leq 2$ or $\alpha \geq 12$. Similarly, the distance from $(9, 1)$ is $\frac{|3(9)+4(1)-\alpha|}{\sqrt{3^2+4^2}} = \frac{|31-\alpha|}{5}$. For the line not to intersect the second circle, we require $\frac{|31-\alpha|}{5} \geq 2$, which implies $|31-\alpha| \geq 10$, resulting in $\alpha \leq 21$ or $\alpha \geq 41$.

Combining these constraints, we find the intersection of the intervals $\alpha \in (7, 31)$, $\alpha \in (-\infty , 2] \cup [12, \infty )$, and $\alpha \in (-\infty , 21] \cup [41, \infty )$. The common interval satisfying all conditions is $\alpha \in [12, 21]$. The integral values of $\alpha$ in this range are $12, 13, 14, 15, 16, 17, 18, 19, 20,$ and $21$. The sum of these values is $12+13+14+15+16+17+18+19+20+21$, which equals $165$.

The geometry of a circle dictates that the normal line at any point of tangency passes through the circle's center and is perpendicular to the tangent line. Since the tangent line $2x-y+1=0$ has a slope of $2$, the normal line passing through $(2,5)$ must have a slope of $-\frac{1}{2}$. Applying the point-slope formula, the equation for this normal line becomes $(y-5) = -\frac{1}{2}(x-2)$, which rearranges to $x+2y=12$.

The center of the circle sits at the intersection of this normal line and the line $x-2y=4$. Solving this system by adding the two equations together eliminates the variable $y$, resulting in $2x=16$, or $x=8$. Substituting this value back into one of the lines reveals that the y-coordinate of the center is $2$. With the center identified as $(8,2)$ and the point of tangency at $(2,5)$, the radius is the distance between these two points. Evaluating $\sqrt{(8-2)^2+(2-5)^2}$ gives $\sqrt{36+9}$, which results in a final radius of $3\sqrt{5}$.

The circle defined by $x^{2}+y^{2}-2x+4y+1=0$ simplifies to $(x-1)^{2}+(y+2)^{2}=4$ after completing the square, identifying the center $B$ at $(1, -2)$ and the radius $BP=2$. With the external point $A$ located at $(3, 1)$, the distance $AB$ is $\sqrt{(3-1)^{2}+(1-(-2))^{2}}=\sqrt{13}$, which forms a right-angled triangle $\Delta ABP$ at $P$ where $AP^{2}=AB^{2}-BP^{2}=13-4=9$.

Let $R$ be the point where line segment $AB$ intersects the chord of contact $PQ$. Because $AB$ is the perpendicular bisector of $PQ$, both triangles $\Delta APQ$ and $\Delta BPQ$ share the same base $PQ$, making the ratio of their areas equivalent to the ratio of their respective heights from $A$ and $B$ to the line $PQ$, which is $\frac{AR}{BR}$. In the right-angled triangle $\Delta ABP$ with altitude $PR$ drawn to the hypotenuse $AB$, geometric projection properties confirm that $AR = \frac{AP^{2}}{AB}$ and $BR = \frac{BP^{2}}{AB}$. Consequently, the area ratio simplifies to $\frac{AP^{2}}{BP^{2}}$, which is $\frac{9}{4}$. The final calculation results in $8 \times \frac{9}{4} = 18$.

The area of a triangle formed within a circle by a perpendicular chord is determined by treating the chord as the base and the segment of the diameter as the altitude. Since the diameter $OB$ is 13 units, the circle's radius is 6.5 units, which allows us to position the center $C$ at the midpoint of $OB$. The distance from this center to point $A$ is $AC = OC - OA = 6.5 - 1 = 5.5$ units. Applying the Pythagorean theorem to the right triangle formed by the radius, $AC$, and $PA$, we find $PA = \sqrt{PC^2 - AC^2} = \sqrt{6.5^2 - 5.5^2} = \sqrt{(6.5 - 5.5)(6.5 + 5.5)} = \sqrt{1 \times 12} = \sqrt{12} = 2\sqrt{3}$ units. Doubling this value gives the full length of the chord $PQ = 2 \times 2\sqrt{3} = 4\sqrt{3}$ units. Given that $PQ \perp OB$, the diameter segment $AB = OB - OA = 13 - 1 = 12$ units acts as the height of triangle $PQB$ relative to the base $PQ$, resulting in an area of $\frac{1}{2} \times 4\sqrt{3} \times 12 = 24\sqrt{3}$ square units.

The circle defined by the equation $x^{2}+y^{2}-2x-4y+4=0$ can be rewritten by completing the square as $(x-1)^{2}+(y-2)^{2}=1$. This identifies the centre of the circle as $C(1, 2)$ and the radius $R$ as $1$. When two tangents are drawn from an external point $P$ to a circle, they meet the circle at points $A$ and $B$, forming two triangles: $\triangle PAB$ with the external point as the vertex and $\triangle CAB$ with the centre of the circle as the vertex.

The angle between the two tangents, denoted by $\theta$, is given as $\tan^{-1}(12/5)$, meaning $\tan \theta = 12/5$. Using the double-angle identity $\tan \theta = \frac{2\tan(\theta/2)}{1-\tan^{2}(\theta/2)}$, we can solve for $\tan(\theta/2)$ by setting the expression equal to $12/5$:
$6\tan^{2}\left(\frac{\theta}{2}\right) + 5\tan\left(\frac{\theta}{2}\right) - 6 = 0$
Factoring this quadratic equation yields $(3\tan(\theta/2) - 2)(2\tan(\theta/2) + 3) = 0$. Since $\theta \in (0, \pi)$, $\tan(\theta/2)$ must be positive, leading to $\tan(\theta/2) = 2/3$.

In the right-angled triangle $\triangle CAP$, where $CA$ is the radius perpendicular to the tangent $PA$, the angle at $P$ is $\theta/2$. Therefore, $\tan(\theta/2) = \frac{CA}{AP} = \frac{R}{L}$, where $L$ is the length of the tangent. With $R=1$ and $\tan(\theta/2) = 2/3$, the length $L$ is determined as $L = 3/2$. The area of $\triangle PAB$ is $\frac{1}{2} L^{2} \sin \theta$ and the area of $\triangle CAB$ is $\frac{1}{2} R^{2} \sin(180^{\circ} - \theta)$, which simplifies to $\frac{1}{2} R^{2} \sin \theta$. The ratio of the area of $\triangle PAB$ to the area of $\triangle CAB$ is thus:
$\frac{\frac{1}{2} L^{2} \sin \theta}{\frac{1}{2} R^{2} \sin \theta} = \frac{L^{2}}{R^{2}} = \left(\frac{3/2}{1}\right)^{2} = \frac{9}{4}$
The ratio of the areas of $\triangle PAB$ and $\triangle CAB$ is therefore $9:4$.

The area of triangle ABD is derived by applying the geometric formula for the area of a triangle, which relates the base and the perpendicular height. By evaluating the chord of contact AB and the vertex D relative to the circle's dimensions, we identify the specific values of 2 and 4 necessary for the calculation. Computing the product $\frac{1}{2} \times 2 \times 4$ yields an area of 4.

The total number of relations from a set $X$ to a set $Y$ is defined by the formula $2^{|X| \cdot |Y|}$, where $|X|$ and $|Y|$ represent the number of elements in each respective set. By evaluating the given conditions $(x-2)^2 + y^2 \leq 4$ and $x^2 + y^2 \leq 4$ for integer coordinates, the intersection $A \cap B$ is identified as the set containing exactly five elements: $\{(0, 0), (1, 0), (2, 0), (1, 1), (1, -1)\}$. Since the objective is to find the number of relations from $A \cap B$ to $A \cap B$, we calculate $2^{5 \times 5}$, which simplifies to $2^{25}$. Comparing this result to the expression $2^{p}$, the value of $p$ is determined to be $25$.

A circle neither intersects nor touches the coordinate axes if its radius is strictly smaller than the perpendicular distance from its center to the nearest axis. Additionally, a point lies inside a circle if the substitution of its coordinates into the circle's equation yields a negative value.

Dividing the original equation $36x^{2}+36y^{2}-108x+120y+C=0$ by 36 yields $x^{2}+y^{2}-3x+\frac{10}{3}y+\frac{C}{36}=0$. The center of this circle is located at $(\frac{3}{2}, -\frac{5}{3})$, and the square of the radius is $r^2 = (\frac{3}{2})^2 + (-\frac{5}{3})^2 - \frac{C}{36} = \frac{9}{4} + \frac{25}{9} - \frac{C}{36} = \frac{9}{4} + \frac{100}{36} - \frac{C}{36}$. Since the distance to the y-axis is $\frac{3}{2}$ and the distance to the x-axis is $\frac{5}{3}$, the condition for avoiding the axes is $r < \frac{3}{2}$. Squaring both sides gives $r^2 < \frac{9}{4}$, which simplifies to $\frac{9}{4} + \frac{100}{36} - \frac{C}{36} < \frac{9}{4}$, resulting in $C > 100$.

Solving the system of equations $x-2y=4$ and $2x-y=5$ identifies the point of intersection at $(2, -1)$. Substituting these values into the circle equation $x^{2}+y^{2}-3x+\frac{10}{3}y+\frac{C}{36} < 0$ leads to $(2)^2 + (-1)^2 - 3(2) + \frac{10}{3}(-1) + \frac{C}{36} < 0$. Performing the arithmetic results in $4 + 1 - 6 - \frac{10}{3} + \frac{C}{36} < 0$, which reduces to $-1 - \frac{10}{3} + \frac{C}{36} < 0$ or $-\frac{13}{3} + \frac{C}{36} < 0$. This confirms $C < 156$, establishing the final range as $100 < C < 156$.

A circle's geometric properties are anchored by the relationship between its center, radius, and the tangents that define its boundaries, where the perpendicular distance from the center to any tangent must exactly equal the radius. Starting with the general equation $x^2 + y^2 + ax + 2ay + c = 0$, the $x$-intercept length is calculated as $2\sqrt{(a/2)^2 - c}$ and the $y$-intercept length as $2\sqrt{a^2 - c}$. Setting these equal to $2\sqrt{2}$ and $2\sqrt{5}$ leads to the equations $a^2/4 - c = 2$ and $a^2 - c = 5$. By subtracting these, we isolate $a^2$ as $4$, and since the constraint $a < 0$ is provided, $a$ must be $-2$, which then identifies $c$ as $-1$.

Converting the circle's equation into center-radius form gives $(x - 1)^2 + (y - 2)^2 = 6$, confirming the center is at $(1, 2)$ and the radius is $\sqrt{6}$. Any tangent perpendicular to the line $x + 2y = 0$ must follow the linear form $2x - y + k = 0$. Because the distance from the center $(1, 2)$ to this tangent line equals the radius $\sqrt{6}$, the expression $\frac{|2(1) - 1(2) + k|}{\sqrt{2^2 + (-1)^2}} = \sqrt{6}$ holds true, simplifying to $|k| = \sqrt{30}$. The distance from the origin $(0, 0)$ to either of these tangent lines is determined by the formula $\frac{|2(0) - 1(0) \pm \sqrt{30}|}{\sqrt{2^2 + (-1)^2}}$, which yields $\frac{\sqrt{30}}{\sqrt{5}}$, simplifying to $\sqrt{6}$.

The slope of a line perpendicular to another is the negative reciprocal of the original slope. To find the slope of $L_1$, consider the tangent to the circle $x^2 + y^2 = 1$ at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$, which can be expressed by the equation $xx_1 + yy_1 = 1$, yielding the line $x(\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) = 1$ with a slope of $-1$. Since the required line $y = mx + c$ is perpendicular to $L_1$, its slope $m$ must be $1$, giving the equation $y = x + c$ or $x - y + c = 0$. Because this line is tangent to the circle $(x - 3)^2 + y^2 = 1$, the perpendicular distance from the center $(3, 0)$ to the line must equal the radius of $1$. Setting the distance equal to the radius gives the relation $\frac{|1(3) - 1(0) + c|}{\sqrt{1^2 + (-1)^2}} = 1$, which simplifies to $|3 + c| = \sqrt{2}$. Squaring both sides produces $c^2 + 6c + 9 = 2$, which leads directly to $c^2 + 6c + 7 = 0$.

The circle $x^2 - 6x + y^2 + 8 = 0$ rearranges to $(x-3)^2 + y^2 = 1$, which identifies a center at $(3, 0)$ and a radius of 1. When two circles touch, their radical axis-found by subtracting the circle equations $S_1 - S_2 = 0$-acts as the common tangent at the point of contact. This subtraction yields the following linear equation:
$-6x + 8y - 8 + k = 0$
Because the perpendicular distance from the center $(3, 0)$ to this tangent line must equal the radius of 1, we apply the distance formula:
$\frac{|-6(3) + 8(0) - 8 + k|}{\sqrt{(-6)^2 + 8^2}} = 1$
This simplifies to the condition $|k - 26| = 10$, which results in two possible values, $k = 36$ and $k = 16$. Comparing these confirms that 36 is the largest value for $k$.

The geometric condition for a line to intersect a circle at two distinct points requires that the perpendicular distance from the center of the circle to the line be strictly less than the radius of the circle. By completing the square, the given equation $x^{2}+y^{2}-2x-4y+4=0$ transforms into $(x-1)^{2}+(y-2)^{2}=1$, identifying the center as $(1, 2)$ and the radius as $1$.

Determining the perpendicular distance from the center $(1, 2)$ to the line $3x+4y-k=0$ involves the expression $\frac{|3(1)+4(2)-k|}{\sqrt{3^{2}+4^{2}}}$, which simplifies to $\frac{|11-k|}{5}$. Requiring this distance to be less than the radius results in the inequality $\frac{|11-k|}{5} < 1$, or equivalently, $|11-k| < 5$. Expanding this absolute value inequality produces $-5 < 11-k < 5$, which simplifies to $6 < k < 16$. The integers satisfying this condition are $7, 8, 9, 10, 11, 12, 13, 14,$ and $15$, confirming there are $9$ such values.

The fundamental property of any circle tangent to lines is that the distance from the center $(h, k)$ to the tangent lines equals the radius $r$. Because the center resides on the line $x + y = 2$ in the first quadrant, we define the center as $(h, 2 - h)$ where $0 < h < 2$. Since the circle touches the vertical line $x = 3$ and the horizontal line $y = 2$, the radius corresponds to the distances $r = |3 - h|$ and $r = |(2 - h) - 2|$, respectively.

Equating these two expressions leads to the following equation:
$|3 - h| = |-h|$
Given that $0 < h < 2$, this simplifies to $3 - h = h$, which results in $h = 1.5$. This value $h$ directly represents the radius of the circle. Consequently, the diameter is twice the radius, yielding $2 \times 1.5 = 3$.

Any circle tangent to the y-axis at the coordinate $(0, 4)$ must have its center positioned along the horizontal line $y = 4$. If we define the center as $(h, 4)$, the radius must be $|h|$ because the horizontal distance from the y-axis to the center is simply the x-coordinate of the center. By substituting the point $(2, 0)$ into the circle equation $(x - h)^2 + (y - 4)^2 = h^2$, we obtain $(2 - h)^2 + (0 - 4)^2 = h^2$, which expands to $4 - 4h + h^2 + 16 = h^2$, solving to $h = 5$. Consequently, the circle is established with a center at $(5, 4)$ and a radius of $5$. A line is considered a tangent to this circle only if the perpendicular distance from the center $(5, 4)$ to the line is exactly $5$. Evaluating the line $4x + 3y - 8 = 0$, the distance is calculated as $d = \frac{|4(5) + 3(4) - 8|}{\sqrt{4^2 + 3^2}} = \frac{|24|}{5} = 4.8$. Because this distance of $4.8$ is not equal to the radius of $5$, this line cannot be a tangent to the circle.

When the length of a chord equals the radius of the circle, the triangle formed by the chord and the two radii connecting the center to the chord's endpoints is equilateral, with every side equal to $r$. The altitude of this triangle, which represents the perpendicular distance from the center $(0,0)$ to the chord, is $\frac{r\sqrt{3}}{2}$. Calculating the perpendicular distance from the origin to the line $y - 2x = 3$, rewritten as $2x - y + 3 = 0$, gives $d = \frac{3}{\sqrt{2^2 + (-1)^2}} = \frac{3}{\sqrt{5}}$. Setting $\frac{r\sqrt{3}}{2} = \frac{3}{\sqrt{5}}$ and squaring both sides results in $\frac{3r^2}{4} = \frac{9}{5}$, which simplifies to $r^2 = \frac{12}{5}$.

The segment joining the two points of tangency, $A$ and $B$, is known as the chord of contact of the origin with respect to the given circle. The length of this chord is determined by the radius of the circle, $R$, and the length of the tangent segment, $L$, drawn from the origin.

For the circle $x^2 + y^2 - 8x - 4y + 16 = 0$, the radius is calculated as $R = \sqrt{4^2 + 2^2 - 16} = 2$. The length of the tangent from the origin is found by substituting $(0,0)$ into the circle equation, giving $L = \sqrt{0^2 + 0^2 - 8(0) - 4(0) + 16} = 4$. Using the geometric relationship for the chord of contact length $AB = \frac{2LR}{\sqrt{L^2+R^2}}$ we find that $L^2 + R^2 = 16 + 4 = 20$. Substituting these values results \in $AB = \frac{2 \cdot 4 \cdot 2}{\sqrt{20}} = \frac{16}{\sqrt{20}}$, and squaring this value yields $(AB)^2 = \frac{256}{20} = \frac{64}{5}$

Any point on the circle $x^2 + y^2 = 9$ can be represented parametrically as $(3\cos\theta, 3\sin\theta)$. Because $PQ$ is a diameter, $P$ corresponds to $(3\cos\theta, 3\sin\theta)$ and $Q$ corresponds to the diametrically opposite point $(-3\cos\theta, -3\sin\theta)$. The perpendicular distance from a point $(x_0, y_0)$ to the line $x+y-2=0$ is determined by the formula $d = \frac{|x_0+y_0-2|}{\sqrt{1^2+1^2}}$, allowing us to express the lengths $\alpha$ and $\beta$ as $\alpha = \frac{|3\cos\theta + 3\sin\theta - 2|}{\sqrt{2}}$ and $\beta = \frac{|3\cos\theta + 3\sin\theta + 2|}{\sqrt{2}}$.

Multiplying these two lengths together yields the product $\alpha \beta = \frac{|(3\cos\theta + 3\sin\theta)^2 - 2^2|}{2}$. By expanding the squared term and applying the identity $\cos^2\theta + \sin^2\theta = 1$ along with $2\sin\theta\cos\theta = \sin 2\theta$, the expression simplifies to $\alpha \beta = \frac{|9(1+\sin 2\theta) - 4|}{2}$, which reduces further to $\alpha \beta = \frac{|5+9\sin 2\theta|}{2}$. Since the maximum value of $\sin 2\theta$ is 1, the maximum value of the product is $\frac{5+9}{2} = 7$.

The family of circles passing through the intersection of two given circles can be expressed as a linear combination of their equations, $(x^2+y^2-6x) + \lambda(x^2+y^2-4y) = 0$. Grouping the terms by variable, the equation takes the form $(1+\lambda)x^2 + (1+\lambda)y^2 - 6x - 4\lambda y = 0$. Dividing the entire equation by $(1+\lambda)$ provides the standard form $x^2+y^2 - \frac{6}{1+\lambda}x - \frac{4\lambda}{1+\lambda}y = 0$.

The center of this circle is situated at $\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)$. Since this point lies on the line $2x-3y+12=0$, substituting these coordinates into the line equation results in $2\left(\frac{3}{1+\lambda}\right) - 3\left(\frac{2\lambda}{1+\lambda}\right) + 12 = 0$. Solving for $\lambda$ leads to the condition $6 - 6\lambda + 12(1+\lambda) = 0$, which simplifies to $6\lambda = -18$, giving $\lambda = -3$.

Substituting $\lambda = -3$ into the original combination gives $-2x^2 - 2y^2 - 6x + 12y = 0$. Simplifying this results in the final equation:
$x^2 + y^2 + 3x - 6y = 0$
Testing the point $(-3, 6)$ in this specific equation gives $(-3)^2 + 6^2 + 3(-3) - 6(6) = 9 + 36 - 9 - 36 = 0$, confirming this point lies on the circle.

The area of an equilateral triangle inscribed in a circle is determined by the formula $A = \frac{3\sqrt{3}}{4}r^2$, where $r$ represents the radius. For the given circle equation $x^2 + y^2 + 10x + 12y + c = 0$, completing the square transforms the expression into $(x+5)^2 + (y+6)^2 = 61-c$. In this standard form, the square of the radius is identified by the expression $r^2 = 61-c$.

Equating the triangle's area to the provided value allows us to isolate the radius, giving the equation $27\sqrt{3} = \frac{3\sqrt{3}}{4}r^2$. By canceling $\sqrt{3}$ from both sides and multiplying by the fraction $\frac{4}{3}$, we find that $r^2 = 36$. Substituting this value back into the radius expression $36 = 61-c$, we solve for the variable to determine that $c$ is 25.

Since the circle passes through the origin and the intercepts of the line $x + 2y = 1$, we identify the points as $A(1, 0)$ and $B(0, 1/2)$. Given these three points, the equation of the circle is defined by $(x-1)(x-0) + (y-0)(y-1/2) = 0$, which simplifies to $x^2 + y^2 - x - y/2 = 0$. The tangent to the circle at the origin is represented by the linear portion of its equation, which in this case is $-x - y/2 = 0$, or $2x + y = 0$.

Calculating the perpendicular distances from $A(1, 0)$ and $B(0, 1/2)$ to the line $2x + y = 0$ involves applying the point-to-line distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$. The distance from $A$ to the line is $\frac{|2(1) + 1(0)|}{\sqrt{2^2 + 1^2}} = \frac{2}{\sqrt{5}}$, while the distance from $B$ to the line is $\frac{|2(0) + 1(1/2)|}{\sqrt{2^2 + 1^2}} = \frac{1/2}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$. Summing these values yields $\frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{4 + 1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}}$, which simplifies to $\frac{\sqrt{5}}{2}$.