Question -
Answer -
Given:
The points (3, 7), (5,5)
The line x тАУ 4y = 1тАж.(1)
By using the standardform of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0 тАж.. (2)
Let us substitute thecentre (-a, -b) in equation (1) we get,
(- a) тАУ 4(- b) = 1
-a + 4b = 1
a тАУ 4b + 1 = 0тАжтАж (3)
Substitute the points(3, 7) in equation (2), we get
32┬а+ 72┬а+2a(3) + 2b(7) + c = 0
9 + 49 + 6a + 14b + c= 0
6a + 14b + c + 58 =0тАж.. (4)
Substitute the points(5, 5) in equation (2), we get
52┬а+ 52┬а+2a(5) + 2b(5) + c = 0
25 + 25 + 10a + 10b +c = 0
10a + 10b + c + 50 =0тАж.. (5)
By simplifyingequations (3), (4) and (5) we get,
a = 3, b = 1, c = тАУ 90
Now, by substitutingthe values of a, b, c in equation (2), we get
x2┬а+ y2┬а+2(3)x + 2(1)y тАУ 90 = 0
x2┬а+ y2┬а+6x + 2y тАУ 90 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬а+ 6x + 2y тАУ 90 = 0.