Question

Find the equation of the circle with (1,3) center which go through the point (2,7)

Answer 1

Answer:

The answer is $x^2+y^2-2x-6y-7=0$

Step-by-step explanation:

Radius of the circle $=$ distance of any point on circle from centre of the circle

                     $=\sqrt{(2-1)^2+(7-3)^2}=\sqrt{1+16}=\sqrt{17}$, using distance formula

Thus using equation of circle with centre $(h,k)$ and radius $r$

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

                     $\Rightarrow (x-1)^2+(y-3)^2=(\sqrt {17})^2\\\Rightarrow x^2+y^2-2x-6y+10=17\\\Rightarrow x^2+y^2-2x-6y-7=0$