Question

Prove that the points $(7,-9)$ and $(11,3)$ lie on a circle with centre at the origin. Also its equation.

Answer 1

Given $(7,-9)d(11,3)$ and center $(0,0)$ calculating distance between the center $(0,0)$ and $(7,-9)$ using distance formula,

$D=\sqrt{(0-7{)}^2+(0+9{)}^2}=\sqrt{49+81}=\sqrt{130}$

Circles equation is $(x-0{)}^2+(y-0{)}^2=\sqrt{130}$

$x^2+y^2=\sqrt{130}$ ------ $\left(1\right)$

Now putting $p+(11,3)$ in the above $e{q}^n\left(1\right)$

$(11{)}^2+(3{)}^2-(\sqrt{130}{)}^2$

$121+9=(\sqrt{130}{)}^2$

$130=130$

Since the equation is true, that point is also on the circle.

$\therefore$ both the points lies on a circle with center at the origin.

Also, the equation is

$x^2+y^2=(\sqrt{130}{)}^2$

$x^2+y^2=130$