Question

Equation of the circle of radius 5 whose centre lies on y-axis in first quadrant and passes through$\left(3,2\right)$ is

• A. $x^2+y^2-12y+11=0$
• B. $x^2+y^2-6y-1=0$
• C. $x^2+y^2-8y+3=0$
• D. None of these

Answer 1

Answer: (A)

$x^2+y^2-12y+11=0$

Let the coordinate of the center be $(0,k)$ as center lies on y-axis

hence, the Equation of circle should be

$(x-0{)}^2+(y-k{)}^2=5^2\Rightarrow x^2+y^2+k^2-2ky=25$

and it passes through the point (3,2)

$3^2+2^2+k^2-2k\cdot 2=25\Rightarrow k^2-4k-12=0\Rightarrow k=-2,6$

But $k=6$ as the center is in the first quadrant

Putting this value in $x^2+y^2+k^2-2ky=25$, we get

$x^2+y^2+36-12y=25\Rightarrow x^2+y^2-12y+11=0$