# How do you find the equation of a circle with center on the line x + 2y= 4 passes through the points A(6,9) and B(12,-9)?

Jan 29, 2018

The equation of circle is ${\left(x - 6\right)}^{2} + {\left(y + 1\right)}^{2} = {10}^{2}$

#### Explanation:

Let $\left(x , y\right)$ be the centre of the circle which passes through points

$A \left(6 , 9\right) \mathmr{and} B \left(12 , - 9\right)$ and $r$ be the radius of the circle.

Distance formula is D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2

$\therefore {r}^{2} = {\left(x - 6\right)}^{2} + {\left(y - 9\right)}^{2} \mathmr{and} {r}^{2} = {\left(x - 12\right)}^{2} + {\left(y + 9\right)}^{2}$

$\therefore {\left(x - 6\right)}^{2} + {\left(y - 9\right)}^{2} = {\left(x - 12\right)}^{2} + {\left(y + 9\right)}^{2}$ or

$\cancel{{x}^{2}} - 12 x + 36 + \cancel{{y}^{2}} - 18 y + \cancel{81} = \cancel{{x}^{2}} - 24 x + 144 + \cancel{{y}^{2}} + 18 y + \cancel{81}$

or $24 x - 12 x - 18 y - 18 y = 144 - 36 \mathmr{and} 12 x - 36 y = 108$ or

$x - 3 y = 9 \left(1\right)$ The centre $\left(x , y\right)$ also satisfy the equation

$x + 2 y = 4 \left(2\right)$ . Solving the equation (1) and equation (2) we

get $x = 6 , y = - 1$ Hence center is at $\left(6 , - 1\right)$ and radius

$\therefore {r}^{2} = {\left(6 - 6\right)}^{2} + {\left(- 1 - 9\right)}^{2} = 100 \therefore r = 10$

The equation of circle is (x – h)^2 + (y – k)^2 = r^2 with the

center being at the point $\left(h = 6 , k = - 1\right)$ and the radius being $r$.

Equation of circle is $\therefore {\left(x - 6\right)}^{2} + {\left(y + 1\right)}^{2} = {10}^{2}$ [Ans]