Question

Find the equation of a circle passing through
the points (1,-4), (5,2) and having its centre
on the line x-2y+9 = 0.​

Answer 1

Question:

Find the equation of a circle passing through the points (1,-4),(5,2) and having its centre is on line x-2y+9=0.

Solution:

Let the Equation of the circle be

${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0...........(1)$

It passes through (1-4)

${(1)}^{2} + ( { - 4)}^{2} + 2g(1) + 2f(1) + c = 0 \\ \\ 1 + 16 + 2g - 8f + c = 0 \\ \\ 17 + 2g - 8f + c = 0 \\ \\ 2g - 8f + c = - 17...........(2)$

Also it passes through (5,2)

$( {5)}^{2} + ( {2)}^{2} + 2g(5) + 2f(2) + c = 0 \\ \\ 25 + 4 + 10g + 4f + c = 0 \\ \\ 29 + 10g + 4f + c = 0 \\ \\ 10g + 4f + c = - 29...............(3).$

Subtract (3) from (4) we get

$10g + 4f + c + 29 - (2g - 8f + c + 17) = 0\\ \\ 10g + 4f + c + 29 - 2g + 8f - c - 17 = 0 \\ \\ 8g + 12f + 12 = 0 \\ \\ 2g + 3f +3 = 0..........(4).$

Centre (-g,-f) lies on x-2y+9=0

$- g + 2f + 9 = 0 \\ \\ g - 2f - 9 = 0.............(5).$

Solve (4) and (5)

$\frac{g}{ - 27 + 6} = \frac{f}{3 + 18} = \frac{1}{ - 4 - 3} \\ \\ \frac{g}{ - 21} = \frac{f}{21} = \frac{1}{ - 7}$

Take

$\frac{g}{ - 21} = \frac{1}{7} \\ \\ g = \frac{ - 21}{7} \\ \\ g = - 3$

And

$\frac{f}{ 21} = \frac{1}{ - 7} \\ \\ f = \frac{21}{ - 7} \\ \\ f = - 3$

Hence the value of g=3 and f=-3

Put value of g , f in (3) we get

$10( - 3) + 4( - 3) + c = - 29 \\ \\ 30 - 12 + c = - 29 \\ \\ c = - 29 - 18 \\ \\ c = - 47$

Now put the value of g, f, c in equation (1)

${x}^{2} + {y}^{2} + 6x - 6y - 47 = 0$