Circles – JEE Main MathematicsPractice Questions & PYQs

The core principle underlying this problem is that a circle is tangent to a line if and only if the perpendicular distance from the center of the circle to that line is exactly equal to the circle's radius. For the given circle $(x-\alpha )^2+(y-\beta )^2=50$, the center is $(\alpha , \beta )$ and the radius is equal to $\sqrt{50}$, which simplifies to $5\sqrt{2}$.

To find the perpendicular distance from the center $(\alpha , \beta )$ to the line $x+y=0$, we use the standard distance formula for a line $Ax+By+C=0$, which gives $\frac{|A\alpha +B\beta +C|}{\sqrt{A^2+B^2}}$. Substituting the known values $A=1$, $B=1$, and $C=0$, this distance becomes $\frac{|\alpha +\beta |}{\sqrt{1^2+1^2}}$, or $\frac{|\alpha +\beta |}{\sqrt{2}}$. Equating this geometric distance to the radius, we establish the equation:

$\frac{|\alpha +\beta |}{\sqrt{2}} = 5\sqrt{2}$

Multiplying both sides by $\sqrt{2}$ results in $|\alpha +\beta | = 10$. Squaring this value directly yields $(\alpha +\beta )^2 = 100$.

For a point $(p, p+1)$ to reside strictly within the defined region $E$, its coordinates must satisfy the inequalities $3-p < p+1$ and $p+1 < \sqrt{9-p^2}$. The first condition, $3-p < p+1$, simplifies to $2 < 2p$, which establishes that $p > 1$. The second condition, $p+1 < \sqrt{9-p^2}$, requires that $p+1$ is positive, which is guaranteed since $p > 1$. Squaring both sides yields $(p+1)^2 < 9-p^2$, which expands to $p^2 + 2p + 1 < 9 - p^2$. Rearranging this inequality gives $2p^2 + 2p - 8 < 0$, or $p^2 + p - 4 < 0$.

Solving the quadratic inequality $p^2 + p - 4 < 0$ determines that $p$ must lie between the roots of $p^2 + p - 4 = 0$, which are $\frac{-1 \pm \sqrt{17}}{2}$. Consequently, the values satisfying this condition fall in the interval $\left(\frac{-1-\sqrt{17}}{2}, \frac{\sqrt{17}-1}{2}\right)$. Intersecting this interval with the initial requirement that $p > 1$ defines the interval $(a, b)$ as $\left(1, \frac{\sqrt{17}-1}{2}\right)$. Identifying the constants from this interval, we have $a = 1$ and $b = \frac{\sqrt{17}-1}{2}$. Since $b$ is a root of the equation $p^2 + p - 4 = 0$, it satisfies the relation $b^2 + b = 4$. Substituting these values into the expression $b^2 + b - a^2$ results in $4 - 1^2$, which equals 3.

The area of triangle $ABC$ formed by sides $2$, $5$ and included angle $60^\circ$ is $\frac{1}{2} \cdot 2 \cdot 5 \cdot \sin(60^\circ) = \frac{5\sqrt{3}}{2} = 2.5\sqrt{3}$. Subtracting this from the total area $4\sqrt{3}$ results in the area of triangle $ADC$ being $1.5\sqrt{3}$. The length of the diagonal $AC$ is determined by the law of cosines as $AC^2 = 2^2 + 5^2 - 2(2)(5)\cos(60^\circ) = 4 + 25 - 10 = 19$. In a cyclic quadrilateral, the opposite angle $\angle D$ is $180^\circ - 60^\circ = 120^\circ$. Utilizing the area formula for triangle $ADC$, $\frac{1}{2} \cdot CD \cdot AD \cdot \sin(120^\circ) = 1.5\sqrt{3}$, which simplifies to $CD \cdot AD = 6$. Applying the law of cosines to triangle $ADC$ yields $AC^2 = CD^2 + AD^2 - 2(CD \cdot AD)\cos(120^\circ)$, leading to $19 = CD^2 + AD^2 + 6$, or $CD^2 + AD^2 = 13$. Given the algebraic identity $(CD+AD)^2 = CD^2 + AD^2 + 2(CD \cdot AD) = 13 + 12 = 25$, it follows that $CD + AD = 5$. The perimeter of the quadrilateral is the sum of all sides $2 + 5 + 5 = 12$.

A circle tangent to the x-axis at $(3, 0)$ with its center in the first quadrant has coordinates $(3, r)$ and a radius of $r$, resulting in the equation $(x - 3)^2 + (y - r)^2 = r^2$. Setting $x=0$ provides the y-intercept, which is determined by the roots of the quadratic $9 + (y - r)^2 = r^2$, simplifying to $y^2 - 2ry + 9 = 0$. The difference between these roots gives the intercept length, $\sqrt{(2r)^2 - 4(9)}$, which equals $6\sqrt{3}$, leading to $4r^2 - 36 = 108$, $r^2 = 36$, and confirming the center is at $(3, 6)$ with a radius of $6$.

The chord length on the line $x - y - 3 = 0$ is calculated using the perpendicular distance $d$ from the center $(3, 6)$ to the line, defined by $d = \frac{|3 - 6 - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|-6|}{\sqrt{2}} = 3\sqrt{2}$. Applying the geometric relation for chord length, $2\sqrt{r^2 - d^2}$, and substituting $r = 6$ and $d = 3\sqrt{2}$ results in $2\sqrt{36 - 18}$. This evaluates to $2\sqrt{18}$, which simplifies to $6\sqrt{2}$.

For a circle with radius $r$ centered such that it cuts equal intercepts from the axes, the relationship between the radius, the perpendicular distance $d$ from the center to the chord $x + y = 1$, and half the chord length $\frac{\sqrt{14}}{2}$ is given by $r^2 = \left(\frac{\sqrt{14}}{2}\right)^2 + d^2$. Using the geometric conditions provided, the distance squared is expressed as $d^2 = \left(r - \frac{1}{\sqrt{2}}\right)^2$. Substituting this into the Pythagorean relation yields $r^2 = \frac{14}{4} + \left(r - \frac{1}{\sqrt{2}}\right)^2$, which simplifies to $r^2 = 3.5 + r^2 - \sqrt{2}r + 0.5$. By canceling $r^2$ from both sides, the equation becomes $\sqrt{2}r = 4$, and solving for the radius gives $r = 2\sqrt{2}$, resulting in $r^2 = 8$.

The geometry of intersecting circles relies on the triangle formed by the centers of the two circles and one of their intersection points. Let the centers be at $O_1(0,0)$ and $O_2(\alpha,0)$, and let $A$ be an intersection point. The sides of $\triangle O_1AO_2$ have lengths 5, 4, and $\alpha$. The common chord is perpendicular to the line connecting the centers, which means the altitude from $A$ to the line $O_1O_2$ is $\frac{\beta}{2}$.

By expressing the area of $\triangle O_1AO_2$ in two different ways, we can relate $\alpha$ and $\beta$ to the given angle $\theta$. Using the base-height formula, the area is $\frac{1}{2} \times \alpha \times \frac{\beta}{2}$, while the side-angle-side area formula provides $\frac{1}{2} \times 5 \times 4 \times \sin \theta$. Equating these expressions leads to $\frac{\alpha \beta}{4} = 10 \sin \theta$. Substituting $\sin \theta = \frac{\sqrt{63}}{8}$ into this equation gives $\alpha \beta = 40 \times \frac{\sqrt{63}}{8}$, which simplifies to $\alpha \beta = 5\sqrt{63}$. Squaring both sides results in $(\alpha \beta)^2 = 25 \times 63 = 1575$.

Since $OC$ is perpendicular to $CP$, the triangle $OCP$ is a right-angled triangle at $C$, which allows us to express the squared distance from the origin to $P$ using the Pythagorean theorem, $OP^2 = OC^2 + CP^2$. First, calculate the distance $OC$ from the origin $(0,0)$ to the center $C(\sqrt{2}, \sqrt{3})$, which gives $OC = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2} = \sqrt{5}$. Given the area of triangle $OCP$ is $\frac{\sqrt{35}}{2}$, we use the formula $\frac{1}{2} \times OC \times CP = \frac{\sqrt{35}}{2}$ to find $CP = \sqrt{7}$. Substituting these values into the Pythagorean relationship yields $OP^2 = 5 + 7 = 12$. Because $Q$ follows the same geometric conditions, $OQ^2$ is also 12, making the sum $a_1^2 + b_1^2 + a_2^2 + b_2^2$ equal to $OP^2 + OQ^2 = 12 + 12 = 24$.