The equation of the circle passing through the point (-1,-3) and touching the line 4 x+3 y-12=0 at the point (3, 0 ) is

Question

The equation of the circle passing through the point (-1,-3) and touching the line 4 x+3 y-12=0 at the point (3, 0 ) is (A) x2+y2-2 x+3 y-3=0 (B) x2+y2+2 x-3 y-5=0 (C) 2 x2+2 y2-2 x+5 y-8=0 (D) None of these

Tardigrade Admin · Accepted Answer

Family of circles touching the line 4 x+3 y-12=0 at point (3,0) is (x-3)2+(y-0)2+λ(4 x+3 y-12)=0 If it passes through the point (-1,-3) then (-1-3)2+(-3-0)2+λ(4(-1)+3(-3)-12)=0 ⇒ λ=1 Hence, equation of circle is (x-3)2+(y-0)2+(4 x+3 y-12) =0 or x2+y2-2 x+3 y-3=0.

Q. The equation of the circle passing through the point $(- 1, - 3)$ and touching the line $4 x + 3 y - 12 = 0$ at the point $(3$ , 0 ) is

$x^{2} + y^{2} - 2 x + 3 y - 3 = 0$

$x^{2} + y^{2} + 2 x - 3 y - 5 = 0$

$2 x^{2} + 2 y^{2} - 2 x + 5 y - 8 = 0$

None of these

Q. The equation of the circle passing through the point (−1,−3) and touching the line 4x+3y−12=0 at the point (3, 0 ) is

x2+y2−2x+3y−3=0

x2+y2+2x−3y−5=0

2x2+2y2−2x+5y−8=0

None of these

Family of circles touching the line 4x+3y−12=0 at point (3,0) is (x−3)2+(y−0)2+λ(4x+3y−12)=0 If it passes through the point (−1,−3) then (−1−3)2+(−3−0)2+λ(4(−1)+3(−3)−12)=0 ⇒λ=1 Hence, equation of circle is (x−3)2+(y−0)2+(4x+3y−12) =0 or x2+y2−2x+3y−3=0.

Q. The equation of the circle passing through the point $(- 1, - 3)$ and touching the line $4 x + 3 y - 12 = 0$ at the point $(3$ , 0 ) is

$x^{2} + y^{2} - 2 x + 3 y - 3 = 0$

$x^{2} + y^{2} + 2 x - 3 y - 5 = 0$

$2 x^{2} + 2 y^{2} - 2 x + 5 y - 8 = 0$