Circles – JEE Main MathematicsPractice Questions & PYQs

To find the required distance, first rewrite the circle equation $x^2 + y^2 - 6x + 8y - 103 = 0$ in the standard form $(x-3)^2 + (y+4)^2 = 128$, identifying the center as $(3, -4)$ and the radius as $8\sqrt{2}$. Because the inscribed square has sides parallel to the coordinate axes, each vertex is located exactly $8$ units horizontally and $8$ units vertically from the center, resulting in vertex coordinates $(3 \pm 8, -4 \pm 8)$. These coordinates identify the four vertices as $(11, 4)$, $(11, -12)$, $(-5, 4)$, and $(-5, -12)$. Calculating the distance of each vertex from the origin $(0, 0)$ gives results of $\sqrt{137}$, $\sqrt{265}$, $\sqrt{41}$, and $13$, from which it is clear that $\sqrt{41}$ is the smallest value.

In any circle, the center of an inscribed rectangle coincides with the center of the circle, which implies that the midpoint of the rectangle's diagonal must lie on any given diameter. Since the rectangle has vertices at $(-8, 5)$ and $(6, 5)$, the other two vertices must be located at $(6, k)$ and $(-8, k)$ to maintain the rectangle's rectangular geometry. The diagonal connecting $(-8, 5)$ and $(6, k)$ therefore has a midpoint at $(-1, \frac{5+k}{2})$.

Because this midpoint must lie on the diameter defined by the line $3y = x + 7$, substituting these coordinates results in $3(\frac{5+k}{2}) = -1 + 7$. Simplifying this equation gives $3(5+k) = 12$, which leads to $5+k = 4$, or $k = -1$. The dimensions of the rectangle are then determined by the horizontal difference, which is $6 - (-8) = 14$, and the vertical difference, which is $5 - (-1) = 6$. Multiplying these side lengths provides the area of $14 \times 6 = 84$.

Two circles intersect at two distinct points if and only if the distance between their centers lies strictly between the absolute difference and the sum of their radii. This principle ensures the circles are close enough to overlap at two points but not so close that one is contained within the other, nor so far apart that they are disjoint.

Identify the geometric properties of the circles from their equations. The first circle, $x^2+y^2 - 16x - 20y + 164 = r^2$, completes the square to become $(x-8)^2 + (y-10)^2 = r^2$, which places the center at $(8, 10)$ with a radius of $r$. The second circle, $(x-4)^2+(y-7)^2 = 36$, has its center at $(4, 7)$ and a radius of $6$. Calculating the distance $d$ between these centers involves finding the difference in coordinates: $d = \sqrt{(8-4)^2 + (10-7)^2} = \sqrt{4^2 + 3^2} = 5$.

Applying the condition $|r - 6| < 5 < r + 6$ establishes the required range for $r$. The inequality $r + 6 > 5$ results in $r > -1$, which is always satisfied for a physical radius. Solving the inequality $|r - 6| < 5$ requires that $r - 6$ sits between $-5$ and $5$, which simplifies to $1 < r < 11$. Therefore, the circles intersect at two distinct points when $r$ falls within this interval.

When a line serves as a boundary between two circles, it must effectively separate them, meaning the two circles lie in opposite half-planes defined by the line. Mathematically, this requires the centers of the circles to yield values with opposite signs when substituted into the linear expression $3x + 4y - \lambda$, and additionally, the perpendicular distance from each center to the line must be at least equal to its corresponding radius to ensure the line does not intersect or pass through either circle.

Simplifying the circle equations reveals that the first circle, $x^2 + y^2 - 2x - 2y + 1 = 0$, has its center at $(1, 1)$ and a radius of $1$, while the second circle, $x^2 + y^2 - 18x - 2y + 78 = 0$, is centered at $(9, 1)$ with a radius of $2$. Substituting these centers into the linear expression $3x + 4y - \lambda$ gives $3(1) + 4(1) - \lambda = 7 - \lambda$ and $3(9) + 4(1) - \lambda = 31 - \lambda$. For the circles to lie on opposite sides, the inequality $(7 - \lambda)(31 - \lambda) < 0$ must hold, which restricts $\lambda$ to the interval $(7, 31)$.

To ensure the line does not cut through the circles, the perpendicular distance from each center to the line $3x + 4y - \lambda = 0$ must satisfy the condition $d \ge r$. For the first circle, this calculation $\left|\frac{7 - \lambda}{5}\right| \ge 1$ results in $\lambda \in (-\infty , 2] \cup [12, \infty )$. For the second circle, the requirement $\left|\frac{31 - \lambda}{5}\right| \ge 2$ yields $\lambda \in (-\infty , 21] \cup [41, \infty )$. Intersecting these two distance-based intervals with the initial separation condition $\lambda \in (7, 31)$ leaves only the values in the range $[12, 21]$.

When two circles intersect at right angles, the radii connecting the centers to the point of intersection serve as the legs of a right triangle. Since the radii are 5 cm and 12 cm, the distance between the two centers acts as the hypotenuse, which calculates to $\sqrt{5^2 + 12^2} = 13$ cm. The common chord of these circles lies perpendicular to this line of centers and represents twice the altitude of this triangle dropped from the intersection point to the hypotenuse.

Calculating the area of this right triangle using the legs gives $\frac{1}{2} \times 5 \times 12 = 30$ cm$^2$. Alternatively, expressing this same area using the hypotenuse as the base and $x$ as the altitude results in the equation $\frac{1}{2} \times 13 \times x = 30$. Solving for $x$ yields $x = \frac{60}{13}$ cm. Because the common chord spans both sides of the center line symmetrically, its total length is $2x$, resulting in $\frac{120}{13}$ cm.

The radical axis of two circles, which corresponds to the line passing through their intersection points $P$ and $Q$, is determined by subtracting one circle equation from the other. Given the circles $x^{2}+y^{2}+5 k x+2 y+k=0$ and $x^{2}+y^{2}+k x+\frac{3y}{2}-\frac{1}{2}=0$, the radical axis is found by subtracting their equations to obtain $4 k x+\frac{1}{2} y+k+\frac{1}{2}=0$.

Since the line $4 x+5 y-k=0$ must pass through $P$ and $Q$, it is identical to this radical axis, implying that the ratios of their corresponding coefficients must be equal. Setting the proportionality $\frac{4k}{4} = \frac{1/2}{5} = \frac{k+1/2}{-k}$ creates a system of conditions for $k$. Solving this relationship results in exactly 2 distinct values of $k$ for which the given line coincides with the radical axis.

When a circle is tangent to a line at a specific point, its equation can be represented as a family of circles in the form $(x-x_1)^2 + (y-y_1)^2 + \lambda L = 0$, where $(x_1, y_1)$ is the point of tangency and $L$ is the equation of the line. For a circle tangent to the line $x - y = 0$ at the point $(1, 1)$, this expression becomes $(x-1)^2 + (y-1)^2 + \lambda(x-y) = 0$. By substituting the coordinates of the point $(1, -3)$ that the circle passes through, we get $(1-1)^2 + (-3-1)^2 + \lambda(1 - (-3)) = 0$. This simplifies to $16 + 4\lambda = 0$, which yields $\lambda = -4$. Substituting this value back into the family equation and expanding terms results in $(x-1)^2 + (y-1)^2 - 4(x-y) = 0$, which simplifies to $x^2 + y^2 - 6x + 2y + 2 = 0$. Comparing this to the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = -3$, $f = 1$, and $c = 2$. Calculating the radius using $\sqrt{g^2 + f^2 - c}$ gives $\sqrt{(-3)^2 + 1^2 - 2}$, which equals $\sqrt{8}$, or $2\sqrt{2}$.

A circle tangent to the x-axis at $(3,0)$ takes the form $(x-3)^2 + (y-f)^2 = f^2$, which expands to $x^{2}+y^{2}-6x-2fy+9=0$. Setting $x=0$ to locate the y-intercepts results in the equation $y^2 - 2fy + 9 = 0$. Because the length of the y-intercept is 8, we set the difference of the roots to 8, expressed as $\sqrt{(2f)^2 - 4(9)} = 8$. Solving this gives $4f^2 - 36 = 64$, which simplifies to $f^2 = 25$, meaning $f = \pm 5$. The resulting circle equation is $x^{2}+y^{2}-6x\pm 10y+9=0$. Testing the point $(3,10)$ in the case $x^{2}+y^{2}-6x-10y+9=0$ yields $9 + 100 - 18 - 100 + 9 = 0$, confirming the point lies on the circle.

The quadrilateral $PC_1QC_2$ is formed by two congruent triangles, $PC_1C_2$ and $QC_1C_2$, which share the line segment $C_1C_2$ as their base. By rearranging the circle equations, the first circle is found to have its center at $C_1(1, 1)$ with a radius of $2$, while the second circle has its center at $C_2(3, 3)$ with a radius of $2$. Calculating the distance between these centers yields $2\sqrt{2}$, and because the sum of the squares of the radii equals the square of the distance between the centers, the triangles are right-angled at the intersection points. Each triangle has an area of $\frac{1}{2} \times 2 \times 2 = 2$, and since the quadrilateral comprises two such congruent triangles, the total area is $2 \times 2 = 4$.

The length of a chord $L$ generated by a line at a perpendicular distance $d$ from the center of a circle with radius $r$ is determined by the Pythagorean relationship $L = 2\sqrt{r^2 - d^2}$. For the circle $x^2 + y^2 = 16$, the radius squared is 16, and the distance from the center to the line $x+y=n$ is $d = \frac{n}{\sqrt{2}}$, resulting in $L^2 = 4(16 - \frac{n^2}{2})$, or $2(32 - n^2)$. Since the chord only exists when the distance is less than the radius, $n \in \mathbb{N}$ must satisfy $\frac{n}{\sqrt{2}} < 4$, which restricts $n$ to 1, 2, 3, 4, 5. Summing these values, we compute $2\sum_{n=1}^{5} (32 - n^2) = 2(31 + 28 + 23 + 16 + 7)$, which equals 210.

The foot of the perpendicular from the origin to the line AB can be denoted as the point $(h, k)$. Since the segment OP is perpendicular to the line AB, the slope of AB must be the negative reciprocal of the slope of OP, expressed as $-\frac{h}{k}$. Applying the point-slope form with the point $(h, k)$, the equation of the line AB becomes $hx + ky = h^2 + k^2$, which locates the axis intercepts at $A$ $(\frac{h^2+k^2}{h}, 0)$ and $B$ $(0, \frac{h^2+k^2}{k})$. Given that the circle passes through the origin and these intercepts, the segment AB represents the circle's diameter with length $2R$. Setting the square of this length to $4R^2$ results in $(\frac{h^2+k^2}{h})^2 + (\frac{h^2+k^2}{k})^2 = 4R^2$, which simplifies to $(h^2+k^2)^2 (\frac{h^2+k^2}{h^2k^2}) = 4R^2$. Multiplying by the denominator confirms the relationship $(h^2+k^2)^3 = 4R^2h^2k^2$, identifying the locus as $(x^2 + y^2)^3 = 4R^2x^2y^2$.

When two circles touch externally, the distance between their centers equals the sum of their radii. The fixed circle has its center at the origin $(0, 0)$ and a radius of $1$, while a moving circle with center $(x, y)$ and radius $r$ that touches the y-axis implies $r = x$. Setting the distance between the two centers equal to the sum of the radii yields $\sqrt{x^2 + y^2} = 1 + x$. Squaring both sides results in $x^2 + y^2 = (1 + x)^2$, which expands to $x^2 + y^2 = 1 + 2x + x^2$. Subtracting $x^2$ from both sides simplifies the relationship to $y^2 = 1 + 2x$, leading to $y = \sqrt{1 + 2x}$ for $x \ge 0$.

When two circles are tangent to one another, their shared tangent line coincides with their radical axis, which is determined by subtracting the equations of the two circles. By subtracting the equation $x^{2}+y^{2}+6x+8y-24=0$ from $x^{2}+y^{2}-4=0$, the quadratic terms cancel out to leave $-6x-8y+20=0$, which simplifies to the linear equation $3x+4y=10$. Testing the provided coordinate options against this equation reveals that the point $(6, -2)$ satisfies the relationship, as $3(6)+4(-2)=18-8=10$.

When three circles of different radii touch each other externally and also rest upon a common tangent line, the distance between the points of tangency on that line for any two circles with radii $r_1$ and $r_2$ is given by $2\sqrt{r_1r_2}$. In a configuration where the smallest circle of radius $a$ sits between two larger circles of radii $b$ and $c$, the total length spanned along the tangent line by the two larger circles is equal to the sum of the distances spanned by each adjacent pair. This geometric arrangement leads to the following equation:

$2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$

Dividing both sides of this equation by $2\sqrt{abc}$ simplifies the relationship, isolating the terms to produce the following expression:

$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$

This inverse relationship mathematically dictates the constraint imposed by the common tangent on circles of these relative sizes.

Any tangent to the unit circle $x^{2}+y^{2}=1$ can be expressed in the form $x \cos \theta + y \sin \theta = 1$ based on its angle of contact. The intercepts on the x and y axes are found by setting the alternate variable to zero, resulting in points $(\frac{1}{\cos \theta}, 0)$ and $(0, \frac{1}{\sin \theta})$. Denoting the midpoint of these intercepts as $(x, y)$, we have $x = \frac{1}{2 \cos \theta}$ and $y = \frac{1}{2 \sin \theta}$, which rearrange to $\cos \theta = \frac{1}{2x}$ and $\sin \theta = \frac{1}{2y}$. Applying the identity $\cos^{2} \theta + \sin^{2} \theta = 1$ leads directly to the equation $(\frac{1}{2x})^{2} + (\frac{1}{2y})^{2} = 1$. Simplifying this expression results in $\frac{1}{4x^{2}} + \frac{1}{4y^{2}} = 1$, which transforms into the locus equation $x^{2} + y^{2} - 4x^{2}y^{2} = 0$.

When two circles touch at a specific point, their common tangent serves as a boundary that allows us to define a family of circles passing through the point of contact. We determine the tangent line to the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ at the point (1, -1), which is $3x - 4y - 7 = 0$. We then express the family of circles as $(x^2 + y^2 + 4x - 6y - 12) + \lambda(3x - 4y - 7) = 0$. Since our circle passes through the point (4, 0), we substitute these coordinates into the family equation to obtain $20 + 5\lambda = 0$, which yields $\lambda = -4$. Applying this value back to the family equation results in $x^2 + y^2 - 8x + 10y + 16 = 0$. The radius is calculated using $\sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + 5^2 - 16}$, which equals 5.

Since the circles share equal radii and intersect at $(0, 1)$ and $(0, -1)$, their centers must lie on the x-axis symmetrically at $(-d, 0)$ and $(d, 0)$. Applying the Pythagorean theorem to the triangle formed by the origin, a center, and an intersection point gives the relation $d^2 + 1^2 = r^2$. The radius connecting the center $(-d, 0)$ to the point $(0, 1)$ has a slope of $1/d$, implying the tangent line at that point has a perpendicular slope of $-d$ and follows the equation $y = -dx + 1$. Because this tangent passes through the second center at $(d, 0)$, substituting these coordinates results in $0 = -d^2 + 1$, confirming $d^2 = 1$, and thus the total distance between the centers is $2d = 2$.

The normal at $(\sqrt{3}, 1)$ to the circle $x^{2}+y^{2}=4$ passes through the origin $(0, 0)$, establishing one vertex at $(0, 0)$ and a side along the x-axis. Using the circle's properties, the tangent line at this point is $\sqrt{3}x + y = 4$, which intersects the x-axis at $x = \frac{4}{\sqrt{3}}$. These lines define a triangle with a base of $\frac{4}{\sqrt{3}}$ along the x-axis and a height of $1$, corresponding to the y-coordinate of the point of tangency. Calculating the area as $\frac{1}{2} \times \text{base} \times \text{height}$ yields $\frac{1}{2} \times \frac{4}{\sqrt{3}} \times 1$, which simplifies to $\frac{2}{\sqrt{3}}$.

The circle $x^2+y^2+2x-4y-4=0$ has center $O_1(-1, 2)$ and radius $r_1 = \sqrt{(-1)^2 + 2^2 - (-4)} = 3$. Since circle $C$ with radius $r=3$ touches $C_1$ externally at $P(2,2)$, the point $P$ is the midpoint of the line segment joining $O_1$ and the center of circle $C$, denoted as $O(h,k)$, because the radii are equal. Equating $P$ to the midpoint of $O_1O$ gives $\frac{h-1}{2} = 2$ and $\frac{k+2}{2} = 2$, which identifies the center of circle $C$ as $O(5,2)$.

The equation of circle $C$ with center $(5,2)$ and radius $3$ is given by
$(x-5)^2 + (y-2)^2 = 9$
Setting $y=0$ to find the x-intercepts yields $(x-5)^2 + 4 = 9$, which simplifies to $(x-5)^2 = 5$. Solving for $x$ gives $x = 5 \pm \sqrt{5}$, and the length of the intercept is the difference between these two points, $|(5+\sqrt{5}) - (5-\sqrt{5})| = 2\sqrt{5}$.

The relationship between the radius $R$ of a circle, the perpendicular distance $d$ from its centre to a chord, and the length of the chord $L$ is given by $R^2 = d^2 + (\frac{L}{2})^2$. The equation of the tangent to the circle $x^2+y^2-2x-1=0$ at the point $(2,1)$ is derived using the transformation $x^2 \to xx_1$, $y^2 \to yy_1$, $x \to \frac{x+x_1}{2}$, and $y \to \frac{y+y_1}{2}$, resulting in $2x + y - (x+2) - 1 = 0$, which simplifies to $x+y-3=0$. The perpendicular distance from the centre $(3,-2)$ of $C_2$ to this tangent line is $d = \frac{|3 + (-2) - 3|}{\sqrt{1^2+1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$. Substituting these values into the radius equation with $L=4$ yields $R^2 = (\sqrt{2})^2 + (\frac{4}{2})^2 = 2+4=6$, resulting in $R = \sqrt{6}$.