Equation of circle

Equation of family of circles:

x2 + y2 - 4x + 3 + k(x - y - 1) = 0

a. For the circle passes through (2 , 0)

(2)2 + 0 - 4(2) + 3 + k(2 - 0 - 1) = 0

k = 1

So, the required equation of the circle:

x2 + y2 - 4x + 3 + (x - y - 1) = 0

x2 + y2 - 3x - y + 2 = 0


b. Rewrite the equation of the family of circles:

x2 + y2 + (k - 4)x - ky + (3 - k) = 0

Radius = √13

i.e. 1/2 √[(k - 4)2 + (-k)2 - 4(3 - k)] = √13

(k2 - 8k + 16) + k2 - 12 + 4k = 4(13)

k2 - 2k - 24 = 0

(k - 6)(k + 4) = 0

k = -4 or 6

Therefore, the equation of circles:

x2 + y2 + (6 - 4)x - 6y + (3 - 6) = 0 or x2 + y2 + (-4 - 4)x - (-4y) + [3 - (-4)] = 0

x2 + y2 + 2x - 6y - 3 = 0 or
x2 + y2 - 8x + 4y + 7 = 0