JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 90

Consider three circles :
$C_1:x^2+y^2=r^2$
$C_2:(x-1)^2+(y-1)^2=r^2$
$C_3:(x-2)^2+(y-1)^2=r^2$
If a line $L:y=mx+c$ be a common tangent to $C_1$ , $C_2$ and $C_3$ such that $C_1$ and $C_3$ lie on one side of line L while $C_2$ lies on other side, then the value of $20\left(r^2+c\right)$ is equal to :

Answer and worked solution

Official answer: D

Line parallel to $C_1C_3$ : $\frac x2-y+c=0$ . Distances: $C_1$ and $C_3$ give $|c|$ , $C_2$ gives $|c-\frac12|$ (all over $\frac{\sqrt5}2$ ).

Opposite sides, equal distances. $c=\frac14$ , $r=\frac1{2\sqrt5}$ , $r^2=\frac1{20}$ .

Value. $20\left(\frac1{20}+\frac14\right)=6$ .

(D)

Topics:Tangents and normals Distance of a point from a line

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JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 14