Question

Find the equation of a circle passing through (9, -23) which is centred at the origin.

Answer 1

Let's first find the radius of the circle. Since it is centred at the origin and passes through (9, -23), the radius will be the distance of (9, -23) from the origin.
=> radius = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\sqrt{(9^2+(-23{)}^2}$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\sqrt{81+529}$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\sqrt{610}$ units

Using the standard form of the equation of a circle,
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $(x-a{)}^2$ + $(y-b{)}^2$ = $r^2$
Here, a = 0, b = 0 and $r^2$ = 610.

Substituting, we get,
$(x-0{)}^2$ + $(y-0{)}^2$ = 610
$x^2$ + $y^2$ = 610