Question

Find the equation of a circle which passes through (2, -3) and (-4, 5) and having the centre on 4x+3y+1= 0​

Answer 1

Answer:

Let the center of the circle (α,β),let radius =r

equation will be x^2+y^2+α^2+β^2-2αx-2βy=r^2

circle passes through (2,-3),(-4,5).

α^2+β^2-4α+6β+13=r^2. (1)

& α^2+β^2+8α-10β+41=r^2. (2)

from (1)&(2)

-12α+16β-28=0

12α-16β=-28. (3)

given centre on line = 4x+3y+1=0

4α+3β=-1

12α+9β=-3. (4)

from (3)&(4)

β=1

β!1 put in (3)

α=-1

by(1)equation

1+1+4+6+13=r^2

r=5

required equation is. x^2+y^2+2x-2y=23.

IF YOU WANT TO CHANGE CENTRE OF CIRCLE

THEN YOU CAN ALSO TAKE CENTRE AS (h,k).

AND JUST REPLACE (α,β) WITH YOUR POINTS

I HOPE THIS HELPS YOU