The equation of a circle with centre (5,4) and touch the y axis is

Question

The equation of a circle with centre (5,4) and touch the y axis is (A) x2+y2-10 x-8 y-16=0 (B) x2+y2-10 x-8 y-61=0 (C) x2+y2+10 x+8 y+16=0 (D) x2+y2-10 x-8 y+16=0

Tardigrade Admin · Accepted Answer

Centre (5,4) circle touches Y-axis So, radius =5 units Equation of circle ⇒ (x-5)2+(y-4)2=52 ⇒ x2+y2-10 x-8 y+16=0

Q. The equation of a circle with centre $(5, 4)$ and touch the $y$ axis is

$x^{2} + y^{2} - 10 x - 8 y - 16 = 0$

$x^{2} + y^{2} - 10 x - 8 y - 61 = 0$

$x^{2} + y^{2} + 10 x + 8 y + 16 = 0$

$x^{2} + y^{2} - 10 x - 8 y + 16 = 0$

Centre $(5, 4)$ circle touches $Y$ -axis So, radius $= 5$ units
Equation of circle $\Rightarrow (x - 5)^{2} + (y - 4)^{2} = 5^{2}$
$\Rightarrow x^{2} + y^{2} - 10 x - 8 y + 16 = 0$

Q. The equation of a circle with centre (5,4) and touch the y axis is

x2+y2−10x−8y−16=0

x2+y2−10x−8y−61=0

x2+y2+10x+8y+16=0

x2+y2−10x−8y+16=0