Question

find the equation of the circle passing through (-2,3) and having centre at (0,0)

Answer 1

answer is obtained as above

Answer 2

Answer:

Concept:
A circle is a collection of all points in a plane that is uniformly distanced from a fixed point. The fixed point is referred to as the circle's center. The radius of a circle is the separation between the center and any point along its circumference.

Step-by-step explanation:

A circle with a (h, k) center and a radius of r has the equation:

             $&(x-h)^{2}+(y-k)^{2}=r^{2}$

This is the equation's standard form. Thus, we can quickly determine the equation of a circle if we know its radius and center coordinates.

Given:

Point  (-2,3)

Center at (0,0)

Find:

The equation of the circle

Solution:

Given center (0, 0) and circle passing through point P(-2,3).

The radius of circle r = distance  "p"

        $r =\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

        $r=\sqrt{(-2-0)^{2}+(3-0)^{2}}$

       $r =\sqrt{4+9}=\sqrt{13}$

Equation of circle having center c(0,0) and Radius $r=\sqrt{13}$ is

        $\begin{aligned}&\text { radious } r=\sqrt{13} \\&(x-h)^{2}+(y-k)^{2}=r^{2} \\&(x-0)^{2}+(y-0)^{2}=(\sqrt{13})^{2} \\&x^{2}+y^{2}=13\end{aligned}$

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