# How to find the standard form of the equation of the specified circle given that it Passes through points A(-1,3) y B (7,-1) and it center is on line 2x+y-11=0?

Aug 8, 2016

${\left(x - 4\right)}^{2} + {\left(y - 3\right)}^{2} = {5}^{2}$

#### Explanation:

Let the coordinate of the center of the circle be $\left(a , b\right) \text{ and its radius be } r$

So the equation of the circle is

$\textcolor{red}{t {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}}$
By the given condition its center lies on the straight line $2 x + y - 11 = 0$

So

$2 a + b = 11. \ldots . . \left(1\right)$

Again $A \left(- 1 , 3\right) \mathmr{and} B \left(7 , - 1\right)$ are on the circle. So

${\left(a + 1\right)}^{2} + {\left(b - 3\right)}^{2} = {\left(a - 7\right)}^{2} + {\left(b + 1\right)}^{2}$

$\implies {\left(a + 1\right)}^{2} + {\left(b - 3\right)}^{2} - {\left(a - 7\right)}^{2} - {\left(b + 1\right)}^{2} = 0$

$\implies 8 \left(2 a - 6\right) - 4 \left(2 b - 2\right) = 0$

$\implies \left(2 a - 6\right) - \left(b - 1\right) = 0$

$\implies 2 a - 6 - b + 1 = 0$

$\implies 2 a - b = 5. \ldots . \left(2\right)$

Subtracting (2) from (1) we get

$2 b = 6 \implies b = 3$

putting the value of b in (2)

$2 a - 3 = 5 \implies a = 4$

$r = \sqrt{{\left(a + 1\right)}^{2} + {\left(b - 3\right)}^{2}}$
$= \sqrt{{\left(4 + 1\right)}^{2} + {\left(3 - 3\right)}^{2}} = 5$
${\left(x - 4\right)}^{2} + {\left(y - 3\right)}^{2} = {5}^{2}$