JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q1

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MathematicsSingle correct+4 / -1Question 1 of 90

Let a circle C touch the lines $L_1:4x-3y+K_1=0$ and $L_2:4x-3y+K_2=0$ , $K_1,K_2\in\mathbb R$ . If a line passing through the centre of the circle C intersects $L_1$ at $(-1,2)$ and $L_2$ at $(3,-6)$ , then the equation of the circle C is :

(x-1)^2+(y-2)^2=4

(x+1)^2+(y-2)^2=4

(x-1)^2+(y+2)^2=16

(x-1)^2+(y-2)^2=16

Answer and worked solution

Official answer: C

Centre. The two tangents are parallel, so the centre is midway between them on any line through it: the midpoint of $(-1,2)$ and $(3,-6)$ , i.e. $(1,-2)$ .

Constants. $L_1$ through $(-1,2)$ : $K_1=10$ . $L_2$ through $(3,-6)$ : $K_2=-30$ .

Radius. Half the distance between the lines: $\dfrac12\cdot\dfrac{|10-(-30)|}5=4$ .

Circle. $(x-1)^2+(y+2)^2=16$ .

(C)

Topics:Equation of a circle Distance of a point from a line

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JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 1