Question

A circle of equation x² + y² + ax + by + c = 0 passes through (0, 6) and touches y = x² at (2, 4). Find a + c.

Answer 1

Correct answer is 16

Given, x²+ y² + ax + by + c = 0 passes through (0, 6)

Then, 0 + 36 + 0 + 6b + c = 0

6b + c = -36 ...(1)

Touches y = x at (2,4)

Thus, passes through (2,4) and also tangent at (2,4) is having same slope.

∴ 4 + 16 + 2a + 4b + c = 0

2a + 4b + c = -20 .... (2)

Now, for slope

dy/dx = 2x

dy/dx_(2.4) = 4

For circle

4b + 32 = -a -4

a + 4b = -36  ....(3)

Putting in (2)

a + a + 4b + c = -20

a + (-36) + c = -20

a + c = 16