How do you write the standard equation for the circle passing through pts (8,-2), (6,2) and (3,-7)?

Dec 31, 2017

The equation of circle is ${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = 25$

Explanation:

Let the points are $P \left(8 , - 2\right) , Q \left(6 , 2\right) , R \left(3 , - 7\right)$ and Centre of

the circle is $\left(a , b\right)$ and radius is $r$

The distance from center to the given 3 points are equal.

${\left(8 - a\right)}^{2} + {\left(- 2 - b\right)}^{2} = {\left(6 - a\right)}^{2} + {\left(2 - b\right)}^{2} = {\left(3 - a\right)}^{2} + {\left(- 7 - b\right)}^{2}$

${\left(8 - a\right)}^{2} + {\left(- 2 - b\right)}^{2} = {\left(6 - a\right)}^{2} + {\left(2 - b\right)}^{2}$

$\therefore 64 - 16 a + {\cancel{a}}^{2} + 4 + 4 b + {\cancel{b}}^{2} = 36 - 12 a + {\cancel{a}}^{2} + 4 - 4 b + {\cancel{b}}^{2}$ or

$- 4 a + 8 b = - 28 \mathmr{and} 2 b - a = - 7 \left(1\right)$ Similarly,

${\left(8 - a\right)}^{2} + {\left(- 2 - b\right)}^{2} = {\left(3 - a\right)}^{2} + {\left(- 7 - b\right)}^{2}$

$\therefore 64 - 16 a + {\cancel{a}}^{2} + 4 + 4 b + {\cancel{b}}^{2} = 9 - 6 a + {\cancel{a}}^{2} + 49 + 14 b + {\cancel{b}}^{2}$ or

$- 10 a - 10 b = = - 10 \mathmr{and} a + b = 1 \left(2\right)$ Adding equation(1) and (2) we

get, $3 b = - 6 \mathmr{and} b = - 2 \therefore a = 1 - b = 1 - \left(- 2\right) = 3 \therefore$

Centre is $\left(3 , - 2\right)$ Radius is $r = | P C | = \sqrt{{\left(8 - a\right)}^{2} + {\left(- 2 - b\right)}^{2}}$

$\therefore r = \sqrt{{\left(8 - 3\right)}^{2} + {\left(- 2 + 2\right)}^{2}} = 5$

The equation of circle is ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ or

${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = 25$ [Ans]