Question

find the coordinates of the centre of a circle and its radius if the equation is 1) x²+y²-10x-12y+6=0

2) -2x²-2y²-18y+9=0

Answer 1

(1)

$x^2+y^2-10x-12y+6=0$

Now by completing the squares

$x^2-10x+(\frac{10}{2})^2+y^2-12y+(\frac{12}{2})^2+6=(\frac{10}{2})^2+(\frac{12}{2})^2\\\implies(x^2-10x+25)+(y^2-12y+36)=2+36-6\\\implies(x-5)^2+(y-6)^2=55..........................(1)$

standard form of the circle is

$(x-h)^2+(y-k)^2=r^2...................................(2)$

where (h,k)is the center and r is radius of the circle by comparing ed. (1) and (2)

we get,

centre(h,k)=(5,6)

$Radius r=\sqrt{55}$

(2)

$-2x^2-2y^2-18y+9=0\\\implies-2(x^2+y^2+9y-\frac{9}{2})=0\\\implies x^2+y^2 +9y-\frac{9}{2}=0$

by completing squares,

$x^2+y^2+9y+(\frac{9}{2})^2=\frac{9}{2}+(\frac{9}{2})^2\\\implies x^2 +y^2+9y+\frac{81}{4}=\frac{9}{2}+\frac{81}{4}\\\implies x^2+(y+\frac{9}{2})^2=\frac{18+81}{4}\\\implies(x-0)^2+(y+(-\frac{9}{2}))^2=\frac{99}{4}$

By comparing from standard form of circle we get,

centre (h,k) =$(0,\frac{-9}{2})$

Radius $r=\sqrt{\frac{99}{4}}=\frac{3}{4}\sqrt11$