Question

Equation of circle which passes through ( 3 , -2) ; ( -2 , 0 ) and center lies on the line y – 2x + 3 = 0

Answer 1

EXPLANATION.

Equation of circle passes through point

( 3,-2) , ( -2,0).

Center lies on the line → y - 2x + 3 = 0.

General equation of circle.

x² + y² + 2gx + 2fy + c = 0.

Circle passes through point (3,-2).

→ (3)² + (-2)² + 2(g)(3) + 2(f)(-2) + c = 0.

→ 9 + 4 + 6g - 4f + c = 0.

→ 13 + 6g - 4f + c = 0. .......(1).

Circle passes through point (-2,0).

→ (-2)² + (0)² + 2(g)(-2) + 2(f)(0) + c = 0.

→ 4 + 0 - 4g + 0 + c = 0.

→ 4 - 4g + c = 0.

→ c = 4g - 4. ........... (2).

Subtract equation (1) and (2) we get,

→ 13 + 6g - 4f + ( 4g - 4 ) = 0.

→ 13 + 6g - 4f + 4g - 4 = 0.

→ 9 + 10g - 4f = 0. ......... (3).

let center ( g, f) lies on line y - 2x + 3 = 0.

→ f - 2g + 3 = 0.

→ f = 2g - 3 ........ (4).

From equation (3) and (4) we get,

→ 9 + 10g - 4(2g - 3 ) = 0.

→ 9 + 10g - 8g + 12 = 0.

→ 2g = 21.

→ g = 21/2.

Put the value of g = 21/2 in equation (4).

→ f = 2(21/2) - 3.

→ f = 21 - 3.

→ f = 18.

From equation (1) we get C.

→ 13 + 6g - 4f + c = 0.

→ 13 + 6(21/2) - 4(18) + c = 0.

→ 13 + 63 - 72 + c = 0.

→ 76 - 72 + c = 0.

→ c = -4.

Equation of circle.

$\sf \: \implies \: (x - h) {}^{2} + (y - k) {}^{2} = {r}^{2} \\ \\ \sf \: \implies \: (x - \frac{21}{2}) {}^{2} + (y - 18) {}^{2} =( { - 4})^{2} \\ \\ \sf \: \implies \: (x - \frac{21}{2}) {}^{2} + (y - 18) {}^{2} = 16$