# How do you write an equation for a circle with point (3, 4) lies on a circle whose centre is at (-1, 2)?

Jul 9, 2016

${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 20$

#### Explanation:

The standard form of the equation of a circle is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (a ,b) are the coordinates of the centre and r, the radius.

We are given the centre but have to find the radius.

Since we are given a point on the circle then the distance from this point to the centre is the $\textcolor{b l u e}{\text{radius}}$

To calculate the radius use the $\textcolor{b l u e}{\text{distance formula}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$

Here the 2 points are (3 ,4) and (-1 ,2)

let $\left({x}_{1} , {y}_{1}\right) = \left(3 , 4\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 1 , 2\right)$

$r = \sqrt{{\left(- 1 - 3\right)}^{2} + {\left(2 - 4\right)}^{2}} = \sqrt{16 + 4} = \sqrt{20}$

Substitute a = -1 , b = 2 and r$= \sqrt{20}$ into the equation of the circle.

${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\sqrt{20}\right)}^{2}$

$\Rightarrow {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 20 \text{ is the equation of the circle}$