# What is the standard form of the equation of a circle given points: (7,-1), (11,-5), (3,-5)?

Jun 21, 2017

Standard form of circle is ${\left(x - 7\right)}^{2} + {\left(y + 5\right)}^{2} = 16$

#### Explanation:

Let the equation of circle be ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$, whose centre is $\left(- g , - f\right)$ and radius is $\sqrt{{g}^{2} + {f}^{2} - c}$. As it passes though $\left(7 , - 1\right)$, $\left(11 , - 5\right)$ and $\left(3 , - 5\right)$, we have

$49 + 1 + 14 g - 2 f + c = 0$ or $14 g - 2 f + c + 50 = 0$ ......(1)

$121 + 25 + 22 g - 10 f + c = 0$ or $22 g - 10 f + c + 146 = 0$ ...(2)

$9 + 25 + 6 g - 10 f + c = 0$ or $6 g - 10 f + c + 34 = 0$ ......(3)

Subtracting (1) from (2) we get

$8 g - 8 f + 96 = 0$ or $g - f = - 12$ ......(A)

and subtracting (3) from (2) we get

$16 g + 112 = 0$ i.e. $g = - 7$

putting this in (A), we have $f = - 7 + 12 = 5$

and putting values of $g$ and $f$ in (3)

$6 \times \left(- 7\right) - 10 \times 5 + c + 34 = 0$ i.e. $- 42 - 50 + c + 34 = 0$ i.e. $c = 58$

andequation of circle is ${x}^{2} + {y}^{2} - 14 x + 10 y + 58 = 0$

and its centre is $\left(7 , - 5\right)$ abd radius is $\sqrt{49 + 25 - 58} = \sqrt{16} = 4$

and standard form of circle is ${\left(x - 7\right)}^{2} + {\left(y + 5\right)}^{2} = 16$

graph{x^2+y^2-14x+10y+58=0 [-3.08, 16.92, -9.6, 0.4]}