A circle has a center that falls on the line y = 8/7x +2 y=87x+2 and passes through ( 7 ,8 )(7,8) and (3 ,9 )(3,9). What is the equation of the circle?

1 Answer
Jul 19, 2016

20x^2+20y^2-189x-296y+1431=020x2+20y2189x296y+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+2gx+2fy+c=0:S:x2+y2+2gx+2fy+c=0.

We know that the Centre CC of SS is C(-g,-f)C(g,f), and, it is given that CC lies on the line :y=8/7x+2:y=87x+2, giving,

-f=-8/7g+2, or, f=8/7g-2....................(1)

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

49+64+14g+16f+c=0.............(2), and,

9+81+6g+18f+c=0..................(3).

We Solve eqns. (1),(2), and, (3).

(2)-(3) rArr 23+8g-2f=0 rArr 2f=8g+23....(4).

By (1), then, 2(8/7g-2)=8g+23, i.e., 16/7g-8g=23+4

:. -40/7g=27 rArr g=-27*7/40=-189/40.

Using (1), f=8/7*(-27*7/40)-2=-27/5-2=-37/5.

Finally, (3) gives, 90-6*189/40-18*37/5+c=0.

:. c=3*189/20+18*37/5-90=567/20+666/5-90.

:. c=(567+2664-1800)/20=1431/20

Hence, S : x^2+y^2-189/20x-74/5y+1431/20=0,

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