The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, -3), then its radius is :

Question

The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, -3), then its radius is : (A) 3 √2  (B) 3 (C) 2 √2  (D) 2

Tardigrade Admin · Accepted Answer

Equation of circle can be written as (x -1)2 + (y -1)2 + l(x -y) = 0 It passes through (1, -3) 16 + λ (4) = 0 ⇒ λ = -4 So (x -1)2 + (y -1)2 -4(x -y) = 0 ⇒ x2 + y2 - 6x + 2y + 2 = 0 ⇒ r = 2 √2

Q. The line $x = y$ touches a circle at the point $(1, 1)$ . If the circle also passes through the point $(1, - 3)$ , then its radius is :

$32$

$3$

$22$

$2$

Equation of circle can be written as $(x - 1)^{2} + (y - 1)^{2} + l (x - y) = 0$
It passes through (1, -3)
$16 + λ (4) = 0 \Rightarrow λ = - 4$
So $(x - 1)^{2} + (y - 1)^{2} - 4 (x - y) = 0$
$\Rightarrow x^{2} + y^{2} - 6 x + 2 y + 2 = 0$
$\Rightarrow r = 22$

Q. The line x=y touches a circle at the point (1,1). If the circle also passes through the point (1,−3), then its radius is :

32​

3

22​

2

Equation of circle can be written as (x−1)2+(y−1)2+l(x−y)=0 It passes through (1, -3) 16+λ(4)=0⇒λ=−4 So (x−1)2+(y−1)2−4(x−y)=0 ⇒x2+y2−6x+2y+2=0 ⇒r=22​

Q. The line $x = y$ touches a circle at the point $(1, 1)$ . If the circle also passes through the point $(1, - 3)$ , then its radius is :

$32$

$3$

$22$

$2$

Equation of circle can be written as $(x - 1)^{2} + (y - 1)^{2} + l (x - y) = 0$
It passes through (1, -3)
$16 + λ (4) = 0 \Rightarrow λ = - 4$
So $(x - 1)^{2} + (y - 1)^{2} - 4 (x - y) = 0$
$\Rightarrow x^{2} + y^{2} - 6 x + 2 y + 2 = 0$
$\Rightarrow r = 22$