Let x 2 + y 2 + A x + B y + C = 0 be a circle passing through (0, 6) and touching the parabola y = x 2 at (2, 4). Then A+C is equal to

Question

Let $x^{2} + y^{2} + A x + B y + C = 0$ be a circle passing through (0, 6) and touching the parabola $y = x^{2}$ at (2, 4). Then A+C is equal to

Smart Achievers · Accepted Answer

(1)

$x^{2} + y^{2} + A x + B y + C = 0$

is passing through $(0, 6) \Rightarrow 6 B + C = - 36$

The tangent of the parabola $y = x^{2}$ at $(2, 4)$ is

$4 x - y - 4 = 0 \dots (1)$

The tangent of circle $x^{2} + y^{2} + A x + B y + C = 0$ at $(2, 4)$ is

$(4 + A) x + (8 + B) y + 2 A + 4 B + 2 C = 0 \dots (2)$

From equation (1) and (2)

$\frac{4 + A}{4} = \frac{8 + B}{- 1} = \frac{2 A + 4 B + 2 C}{- 4}$

$A + 4 B = - 36 \dots (3)$

$3 A + 4 B + 2 C = - 4 \dots (4)$

From equation (3) and (4) $A + C = 16$

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