A circle has a center that falls on the line y = 7/4x +4  and passes through  ( 4 ,7 ) and (7 ,5 ). What is the equation of the circle?

Jun 28, 2016

$2 {x}^{2} + 2 {y}^{2} + 100 x + 159 y - 1643 = 0$

Explanation:

General equation of circle can be represented as ${x}^{2} + {y}^{2} - 2 h x - 2 k y + c = 0$ -> $E q u a t i o n 1$

where $\left(h , k\right)$ represents the center of the circle , $r$- radius of the circle, $c = {h}^{2} + {k}^{2} - {r}^{2}$

As we know the given points $\left(4 , 7\right) \mathmr{and} \left(7 , 5\right)$ lie on the circle, they must satisfy Equation 1.

Putting the given points $\left(4 , 7\right) \mathmr{and} \left(7 , 5\right)$ in the above equation of circle (1) , we get :

${4}^{2} + {7}^{2} - 2 h \cdot 4 - 2 k \cdot 7 + c = 0$

=>$- 8 h - 14 k + c = - 65$ =>

$8 h + 14 k - c = 65$ -> $E q u a t i o n 2$

${7}^{2} + {5}^{2} - 2 h \cdot 7 - 2 k \cdot 5 + c = 0$

=>$- 14 h - 10 k + c = - 74$ =>

$14 h + 10 k - c = 74$ -> $E q u a t i o n 3$

Since the centre of the circle is $\left(h , k\right)$ lies on the line $y = \frac{7}{4} x + 4$, $\left(h , k\right)$ must satisfy the equation of line.

$k = \frac{7}{4} \cdot \left(h\right) + 4$ =>

$7 h - 4 k = - 16$ -> $E q u a t i o n 4$

Subtract Equation 2 and 3

=> $\left(14 h - 8 h\right) + \left(10 k - 14 k\right) + \left(- c + c\right) =$ 74 - 65

$6 h - 4 k = 9$ -> $E q u a t i o n 5$

=>$\left(- 7 h + 6 h\right) + \left(4 k - 4 k\right) = \left(16 + 9\right)$ -> $E q u a t i o n 5$

$- h = 25 \implies h = - 25$

-4k= 9 - (6*(-25)#

$k = - \frac{159}{4}$

Solving for c in Equation 3, we get $c = 14 \cdot \left(- 25\right) + 10 \cdot \left(- \frac{159}{4}\right) - 74$

$c = - \frac{3286}{4}$ = $- \frac{1643}{2}$

Circle equation is:

${x}^{2} + {y}^{2} - 2 \left(- 25\right) x - 2 \left(- \frac{159}{4}\right) y - \frac{1643}{2} = 0$

${x}^{2} + {y}^{2} + 50 x - \left(- \frac{159}{2}\right) y - \frac{1643}{2} = 0$

$2 {x}^{2} + 2 {y}^{2} + 100 x + 159 y - 1643 = 0$