A circle C touches the line x=2 y at the point (2,1) and intersects the circle C1: x2+y2+2 y-5=0 at two points P and Q such that PQ is a diameter of C1 . Then the diameter of C is :

Question

A circle C touches the line x=2 y at the point (2,1) and intersects the circle C1: x2+y2+2 y-5=0 at two points P and Q such that PQ is a diameter of C1 . Then the diameter of C is : (A) 7 √5 (B) 15 (C) √285 (D) 4 √15

Tardigrade Admin · Accepted Answer

(x-2)2+(y-1)2+λ(x-2 y)=0 C: x2+y2+x(λ-4)+y(-2-2 λ)+5=0 C1: x2+y2+2 y-5=0 S1-S2=0 (Equation of PQ) (λ-4) x-(2 λ+4) y+10=0 Passes through (0,-1) ⇒ λ=-7 C: x2+y2-11 x+12 y+5=0 =(√245/4) Diometer =7 √5

Q. A circle $C$ touches the line $x = 2 y$ at the point $(2, 1)$ and intersects the circle $C_{1} : x^{2} + y^{2} + 2 y - 5 = 0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1} .$ Then the diameter of $C$ is :

$75$

$15$

$285$

$415$

Q. A circle C touches the line x=2y at the point (2,1) and intersects the circle C1​:x2+y2+2y−5=0 at two points P and Q such that PQ is a diameter of C1​. Then the diameter of C is :

75​

15

285​

415​

(x−2)2+(y−1)2+λ(x−2y)=0 C:x2+y2+x(λ−4)+y(−2−2λ)+5=0 C1​:x2+y2+2y−5=0 S1​−S2​=0 (Equation of PQ) (λ−4)x−(2λ+4)y+10=0 Passes through (0,−1) ⇒λ=−7 C:x2+y2−11x+12y+5=0 =4245​​ Diometer =75​

Q. A circle $C$ touches the line $x = 2 y$ at the point $(2, 1)$ and intersects the circle $C_{1} : x^{2} + y^{2} + 2 y - 5 = 0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1} .$ Then the diameter of $C$ is :

$75$

$15$

$285$

$415$