A circle has a center that falls on the line y = 8/7x +2  and passes through  ( 7 ,8 ) and (3 ,9 ). What is the equation of the circle?

Jul 19, 2016

$20 {x}^{2} + 20 {y}^{2} - 189 x - 296 y + 1431 = 0$. graph{20x^2+20y^2-189x-296y+1431=0 [-10, 10, -5, 5]}

Explanation:

Suppose that eqn. of circle is, $: S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$.

We know that the Centre $C$ of $S$ is $C \left(- g , - f\right)$, and, it is given that $C$ lies on the line $: y = \frac{8}{7} x + 2$, giving,

$- f = - \frac{8}{7} g + 2 , \mathmr{and} , f = \frac{8}{7} g - 2. \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

We are also given that the pts.$\left(7 , 8\right) \mathmr{and} \left(3 , 9\right)$ lie on $S$. Hence, these co-ords. must satisfy the eqn. of $S$. Accordingly,

$49 + 64 + 14 g + 16 f + c = 0. \ldots \ldots \ldots \ldots \left(2\right)$, and,

$9 + 81 + 6 g + 18 f + c = 0. \ldots \ldots \ldots \ldots \ldots . . \left(3\right)$.

We Solve eqns. $\left(1\right) , \left(2\right) , \mathmr{and} , \left(3\right)$.

$\left(2\right) - \left(3\right) \Rightarrow 23 + 8 g - 2 f = 0 \Rightarrow 2 f = 8 g + 23. \ldots \left(4\right)$.

By $\left(1\right)$, then, $2 \left(\frac{8}{7} g - 2\right) = 8 g + 23 , i . e . , \frac{16}{7} g - 8 g = 23 + 4$

$\therefore - \frac{40}{7} g = 27 \Rightarrow g = - 27 \cdot \frac{7}{40} = - \frac{189}{40}$.

Using $\left(1\right) , f = \frac{8}{7} \cdot \left(- 27 \cdot \frac{7}{40}\right) - 2 = - \frac{27}{5} - 2 = - \frac{37}{5}$.

Finally, $\left(3\right)$ gives, $90 - 6 \cdot \frac{189}{40} - 18 \cdot \frac{37}{5} + c = 0$.

$\therefore c = 3 \cdot \frac{189}{20} + 18 \cdot \frac{37}{5} - 90 = \frac{567}{20} + \frac{666}{5} - 90$.

$\therefore c = \frac{567 + 2664 - 1800}{20} = \frac{1431}{20}$

Hence, $S : {x}^{2} + {y}^{2} - \frac{189}{20} x - \frac{74}{5} y + \frac{1431}{20} = 0$,

$S : 20 {x}^{2} + 20 {y}^{2} - 189 x - 296 y + 1431 = 0$.