Circles – JEE Main MathematicsPractice Questions & PYQs

Let the center of the circle be $(h, k)$. Since this point lies on the line $y - 4x + 3 = 0$, it satisfies the condition $k = 4h - 3$. Because the points $(2, 3)$ and $(4, 5)$ lie on the circle, their squared distances to the center must be equal:
$(h-2)^2 + (k-3)^2 = (h-4)^2 + (k-5)^2$
Expanding both sides leads to $h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 8h + 16 + k^2 - 10k + 25$, which simplifies to $4h + 4k = 28$ or $h + k = 7$. Substituting $k = 4h - 3$ into this linear equation gives $h + 4h - 3 = 7$, resulting in $5h = 10$, so $h = 2$ and $k = 5$. The radius is calculated using the distance from the center $(2, 5)$ to the point $(2, 3)$ as $r = \sqrt{(2-2)^2 + (5-3)^2} = \sqrt{4} = 2$.

A line is tangent to a circle if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle. We first find the tangent equation to the curve $x^2 = y - 6$ at $(1, 7)$ by calculating the slope $\frac{dy}{dx} = 2x$, which equals $2$ at $x = 1$, resulting in the linear equation $y - 7 = 2(x - 1)$, or $2x - y + 5 = 0$.

For the circle $x^2 + y^2 + 16x + 12y + c = 0$, completing the square identifies the center as $(-8, -6)$ and the radius as $\sqrt{(-8)^2 + (-6)^2 - c} = \sqrt{100 - c}$. Applying the condition for tangency, the distance from $(-8, -6)$ to the line $2x - y + 5 = 0$ is $\frac{|2(-8) - (-6) + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|-5|}{\sqrt{5}} = \sqrt{5}$. Equating this distance to the radius $\sqrt{100 - c}$ and squaring both sides gives $100 - c = 5$, which leads to $c = 95$.

The product of the lengths of the segments formed by a secant line passing through a point $P(x_1, y_1)$ and intersecting a circle $x^2 + y^2 - r^2 = 0$ at points $A$ and $B$ is equal to the power of the point with respect to the circle, defined by the formula $PA \cdot PB = x_1^2 + y_1^2 - r^2$. Substituting the coordinates of the given point $P(4, 7)$ and the circle equation $x^2 + y^2 - 9 = 0$ into this formula gives $PA \cdot PB = 4^2 + 7^2 - 9$. Calculating the values results in $16 + 49 - 9 = 56$.

Because both the parabola $y = 4 - x^2$ and the lines $y = |x|$ exhibit symmetry about the $y$-axis, the center of the circle must lie on the $y$-axis. By positioning the circle to touch the parabola at its vertex $(0, 4)$, we establish the coordinates of the center as $(0, 4 - r)$, where $r$ denotes the radius.

The requirement that the circle remains tangent to the lines $y = |x|$ implies that the perpendicular distance from the center $(0, 4 - r)$ to the line $x - y = 0$ must equal the radius $r$. Setting up this distance using the standard formula gives $\frac{|0 - (4 - r)|}{\sqrt{1^2 + (-1)^2}} = r$, which simplifies to the expression $|4 - r| = r\sqrt{2}$. Solving for $r$ under the condition that $r < 4$, we obtain the equation $4 - r = r\sqrt{2}$, which rearranges to $r(1 + \sqrt{2}) = 4$. Multiplying by the conjugate of the denominator results in $4(\sqrt{2} - 1)$.

The maximum squared distance between a fixed point $P$ and any point $Q$ on a circle occurs when $Q$ lies on the line passing through $P$ and the circle's center $C$, specifically at the point furthest from $P$. This maximum length is $d+r$, where $d$ is the distance from $P$ to the center $C$ and $r$ is the circle's radius; thus, the maximum value of $(PQ)^2$ is $(d+r)^2$. Converting the circle equation $x^2 + y^2 - 5x - y + 5 = 0$ into standard form by completing the square yields the center $C = (\frac{5}{2}, \frac{1}{2})$ and the radius squared $r^2 = \frac{3}{2}$.

Calculating the distance $d$ between $P(0, -2)$ and $C(\frac{5}{2}, \frac{1}{2})$ involves the distance formula, resulting in $d^2 = (\frac{5}{2}-0)^2 + (\frac{1}{2}-(-2))^2 = \frac{25}{4} + \frac{25}{4} = \frac{25}{2}$. Using the expansion $(d+r)^2 = d^2 + r^2 + 2dr$, the first part is $d^2 + r^2 = \frac{25}{2} + \frac{3}{2} = 14$. The cross term $2dr$ becomes $2 \cdot \sqrt{\frac{25}{2}} \cdot \sqrt{\frac{3}{2}} = 2 \cdot \frac{5}{\sqrt{2}} \cdot \frac{\sqrt{3}}{\sqrt{2}} = 2 \cdot \frac{5\sqrt{3}}{2} = 5\sqrt{3}$, meaning the final maximum value is $14+5\sqrt{3}$.

Calculating the distance from the center to a chord involves constructing a right-angled triangle where the radius serves as the hypotenuse and the perpendicular distance acts as the adjacent side to an angle that is half the size of the central angle. Given the circle diameter is 4, the radius is 2, and because the two central angles are $\cos^{-1}\left(\frac{1}{7}\right)$ and $\sec^{-1}(7)$, both chords subtend the same angle since $\sec^{-1}(7)$ is equivalent to $\cos^{-1}\left(\frac{1}{7}\right)$. Using the half-angle formula $d = 2 \cos\left(\frac{\theta}{2}\right)$ with $\theta = \cos^{-1}\left(\frac{1}{7}\right)$, the calculation $\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}}$ leads to $\sqrt{\frac{1 + 1/7}{2}} = \sqrt{\frac{8/7}{2}} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$. Consequently, the perpendicular distance to each chord is $2 \times \frac{2}{\sqrt{7}} = \frac{4}{\sqrt{7}}$. Since the chords lie on opposite sides of the center, the total separation between them is the sum of these two distances, which results in $\frac{4}{\sqrt{7}} + \frac{4}{\sqrt{7}} = \frac{8}{\sqrt{7}}$.

The geometry of a circle dictates that the tangent line at any point is perpendicular to the radius connecting that point to the center. Solving the system of linear equations $x - y = 1$ and $2x + y = 3$ provides the circle's center; adding these yields $3x = 4$, so the x-coordinate is $4/3$ and the y-coordinate is $1/3$. Calculating the slope of the radius segment between the center $(4/3, 1/3)$ and the point of tangency $(1, -1)$ involves finding the ratio of the difference in y-coordinates to the difference in x-coordinates, which is $m = \frac{1/3 - (-1)}{4/3 - 1} = 4$. Since the tangent line must be perpendicular to this radius, its slope is the negative reciprocal, $m' = -1/4$. Using the point-slope form $y - y_1 = m'(x - x_1)$ with the contact point $(1, -1)$, we arrive at the expression $y - (-1) = -1/4(x - 1)$. Simplifying this yields the following:
$4(y + 1) = -(x - 1)$
Rearranging the terms results in $x + 4y + 3 = 0$.

Since the circle is tangent to the y-axis at $(0, 2)$, the family of such circles is defined by the equation $(x-0)^2 + (y-2)^2 + \lambda x = 0$. Substituting the point $(-2, 4)$ into this expression allows for the determination of $\lambda$, as $(-2)^2 + (4-2)^2 + \lambda(-2) = 0$ simplifies to $4 + 4 - 2\lambda = 0$, revealing that $\lambda = 4$. This leads to the specific circle equation
$x^2 + y^2 + 4x - 4y + 4 = 0$
The center of this circle is found at $(-2, 2)$ by inspecting the coefficients of the quadratic equation. Because a diameter must necessarily pass through the center of the circle, substituting the coordinates $(-2, 2)$ into the provided options confirms that the equation $2x - 3y + 10 = 0$ is the valid choice, as $2(-2) - 3(2) + 10$ equals zero.

The radius of circle $S$ is derived from the geometric relationship between a circle's chord length, the perpendicular distance from its center to that chord, and the circle's radius. The given equation $x^2 + y^2 - 4x + 6y - 12 = 0$ rearranges to $(x-2)^2 + (y+3)^2 = 25$, identifying a center at $(2, -3)$ and a radius of 5. Because a diameter of this circle serves as a chord for circle $S$, the chord's total length equals the diameter, 10, and its midpoint coincides with the center $(2, -3)$.

The perpendicular distance $d$ from the center of circle $S$ at $(-3, 2)$ to this chord is simply the distance between the two centers, calculated as $\sqrt{(2 - (-3))^2 + (-3 - 2)^2} = \sqrt{25 + 25} = \sqrt{50}$. Using the Pythagorean theorem for circle $S$, where the radius $R$, the chord's half-length $a=5$, and the distance $d$ form a right triangle, we have $R^2 = a^2 + d^2$. Substituting the values yields $R^2 = 25 + 50 = 75$, which results in a final radius of $5\sqrt{3}$.

A parabola is geometrically defined as the locus of points that are equidistant from a fixed point, known as the focus, and a fixed line, known as the directrix. When a circle touches both a given fixed circle and a fixed line, the locus of its center satisfies this equidistant property, which characterizes a parabolic path. For the fixed circle defined by $x^2+y^2-8x-8y-4=0$, we can determine its center and radius by rewriting the equation as $(x-4)^2 + (y-4)^2 = 36$, identifying the center as $(4, 4)$ and the radius as $6$.

The moving circle with center $(x, y)$ is tangent to the x-axis, meaning its radius is equal to its perpendicular distance from the x-axis, given by $|y|$. Because this circle touches the fixed circle externally, the distance between the two centers must be equal to the sum of their radii, resulting in the relation $\sqrt{(x-4)^2 + (y-4)^2} = |y| + 6$. Squaring both sides of this equation yields $(x-4)^2 + (y-4)^2 = (|y| + 6)^2$. Expanding this expression and simplifying, specifically for values where $y \ge 0$, leads to $(x-4)^2 = 20y + 20$. Since this equation takes the standard form of a parabola, the collection of all such centers forms a parabolic curve.

The family of lines given by $(2x-3y+4)+k(x-2y+3)=0$ represents a collection of lines that all pass through the fixed intersection point of $2x-3y+4=0$ and $x-2y+3=0$. Solving these two equations simultaneously identifies this common intersection point at $(1, 2)$.

Reflecting any point $P(2, 3)$ across a line that passes through a fixed point $A(1, 2)$ preserves the distance between the original point and the intersection point, meaning the distance $AP$ must equal the distance from $A$ to the image point $P'(x, y)$. This constant distance is calculated as $AP = \sqrt{(2-1)^2 + (3-2)^2} = \sqrt{2}$. Since the image point maintains a fixed distance of $\sqrt{2}$ from the point $(1, 2)$, the locus of the image points is a circle with radius $\sqrt{2}$.

The number of common tangents between two circles depends entirely on the relationship between the distance separating their centers and the sum of their radii. By identifying the parameters from the given equations, we find the first circle has center $C_1(2,3)$ and radius $r_1=5$, while the second has center $C_2(-3,9)$ and radius $r_2=8$. The distance between these two centers is $C_1C_2=\sqrt{25+144}=13$. Because this calculated distance is exactly equal to the sum of the radii $r_1+r_2=13$, the circles touch each other externally, which necessitates the existence of exactly three common tangents.

When two circles touch externally, the distance between their centers is exactly equal to the sum of their radii. Circle C has its center at $(1,1)$ and a radius of $1$, while circle T, centered at $(0,y)$ and passing through the origin, has a radius equal to the distance from $(0,y)$ to $(0,0)$, which is $y$ for $y>0$. Equating the distance between the center of circle C $(1,1)$ and the center of circle T $(0,y)$ to the sum of the radii $1+y$ leads to the equation:

$\sqrt{1^2 + (1-y)^2} = 1+y$

Squaring both sides gives $1 + (1-y)^2 = (1+y)^2$, which expands to $1 + 1 - 2y + y^2 = 1 + 2y + y^2$. This expression simplifies further to $2 - 2y = 1 + 2y$, and solving for $y$ yields $4y = 1$, confirming the radius of T is $1/4$.

A circle that touches the x-axis at the point $(3, 0)$ has a center with an x-coordinate of 3 and a radius equal to the absolute value of the center's y-coordinate, allowing the equation to be expressed as $(x-3)^2 + y^2 + \lambda y = 0$. By substituting the known point $(1, -2)$ through which the circle passes, we obtain $(1-3)^2 + (-2)^2 + \lambda(-2) = 0$, which simplifies to $4 + 4 - 2\lambda = 0$. Solving this yields $\lambda = 4$, completing the circle equation as $(x-3)^2 + y^2 + 4y = 0$. To determine which point lies on this circle, we test the coordinates provided. Substituting $(5, -2)$ into the equation results in $(5-3)^2 + (-2)^2 + 4(-2) = 4 + 4 - 8$, which equals zero and confirms that this point satisfies the equation.

When a circle touches the $x$-axis at $(1, 0)$, its center must have an $x$-coordinate of 1 and a $y$-coordinate, $h$, that equals the radius. Equating the distance from the center $(1, h)$ to the point $(2, 3)$ with the radius $h$ leads to the equation:

$(1 - 2)^2 + (h - 3)^2 = h^2$

Expanding this results in:

$1 + h^2 - 6h + 9 = h^2$

This simplifies to $6h = 10$, which results in $h = \frac{5}{3}$. Since the diameter is twice the radius, the final length is $2h = \frac{10}{3}$.

Two circles touch internally when the distance between their centers equals the absolute difference of their radii. For the first circle given by $x^2+y^2=ax$, completing the square gives the form $(x - a/2)^2 + y^2 = (a/2)^2$, identifying its center at $(a/2, 0)$ and a radius of $r_1 = |a/2|$. The second circle $x^2+y^2=c^2$ has its center at $(0, 0)$ and a radius $r_2 = c$. The distance between these two centers is $d = |a/2 - 0| = |a/2|$. Setting this distance equal to the difference of the radii requires that $|a/2| = | |a/2| - c |$. Solving this equation leads to $c = |a|$, since the other possible result, $c=0$, is excluded by the problem constraints.

The geometric condition for a straight line to intersect a circle at two distinct points is that the perpendicular distance from the circle's center to the line must be strictly less than the radius. First, we identify the properties of the circle from the equation $x^2 + y^2 - 4x - 8y - 5 = 0$, which reveals a center at $(2, 4)$ and a radius of $r = \sqrt{2^2 + 4^2 + 5} = 5$.

For the line $3x - 4y - m = 0$ to cut through the circle, the perpendicular distance $d$ from the center to the line must satisfy $d < r$. Substituting the center coordinates $(2, 4)$ into the distance formula gives the following inequality:
$\frac{|3(2) - 4(4) - m|}{\sqrt{3^2 + (-4)^2}} < 5$
Simplifying the terms within the expression results in $\frac{|6 - 16 - m|}{5} < 5$, which leads to $|-10 - m| < 25$. Because the absolute value of a number is the same as the absolute value of its negative, this is equivalent to $|10 + m| < 25$. Expanding this into the inequality $-25 < 10 + m < 25$ and subtracting 10 from all parts confirms the required range is $-35 < m < 15$.

The family of circles passing through the intersection points $P$ and $Q$ of two given circles can be expressed using the equation $S_1 + \lambda L = 0$, where $S_1 = 0$ is the equation of one circle and $L = S_1 - S_2 = 0$ is the equation of the common chord. Subtracting the second circle equation from the first, the common chord is found to be $x + 5y + p^2 + 2p - 5 = 0$. Substituting the first circle equation and the common chord into the family formula gives $(x^2+y^2+3x+7y+2p-5) + \lambda(x + 5y + p^2 + 2p - 5) = 0$.

For this circle to pass through the point $(1,1)$, we substitute $x=1$ and $y=1$ into the expression, resulting in $(1+1+3+7+2p-5) + \lambda(1+5+p^2+2p-5) = 0$. Simplifying this leads to $(2p+7) + \lambda(p^2+2p+1) = 0$, which can be rearranged to find the required parameter $\lambda = -\frac{2p+7}{(p+1)^2}$. A valid circle exists for any finite value of $\lambda$, meaning the construction is possible for all values of $p$ except where the denominator becomes zero. Since $(p+1)^2 = 0$ only at $p=-1$, a circle exists for all values of $p$ except this single case.

Any point that maintains a constant ratio of distances from two fixed points in a plane defines a Circle of Apollonius. Since the three distinct points $A$, $B$, and $C$ all satisfy the condition that their distances from $P(1, 0)$ and $Q(-1, 0)$ are in the fixed ratio of $\frac{1}{3}$, they must all lie on the circumference of this specific circle. Consequently, the circumcenter of the triangle formed by these three points is identical to the center of this circle.

To derive the equation of this circle, consider an arbitrary point $(x, y)$ on the locus such that the ratio $\frac{AP}{AQ} = \frac{1}{3}$ holds true. Squaring both sides of $3AP = AQ$ gives $9AP^2 = AQ^2$, which translates to $9((x-1)^2 + y^2) = (x+1)^2 + y^2$ when using the distance formula. Expanding these terms results in $9(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2$, which simplifies to $8x^2 - 20x + 8y^2 + 8 = 0$. Dividing the entire equation by $8$ provides the standard form $x^2 - \frac{5}{2}x + y^2 + 1 = 0$. The center of a circle defined by $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $(-g, -f)$, which, for this equation, confirms the center is located at $\left(\frac{5}{4}, 0\right)$.

The center of a circle is the midpoint of any diameter, implying that the average of the coordinates of two diametrically opposite points must equal the center of the circle. For the circle $x^2+y^2+2x+4y-3=0$, identifying the coefficients of $x$ and $y$ reveals a center of $(-1,-2)$.

Let the required point be $Q(\alpha, \beta)$. By setting the midpoint of the given point $P(1,0)$ and the unknown point $Q(\alpha, \beta)$ equal to the center $(-1, -2)$, we establish the following equations:
$\frac{1+\alpha}{2} = -1 \quad \text{and} \quad \frac{0+\beta}{2} = -2$
Solving these equations results in $\alpha = -3$ and $\beta = -4$, identifying the diametrically opposite point as $(-3,-4)$.