Tangent to the curve y = x 2 + 6 at a point (1, 7) touches the circle x 2 + y 2 + 16 x + 12 y + c = 0 at a point Q . Then the coordinates of Q are [2005]

Question

Tangent to the curve $y = x^{2} + 6$ at a point (1, 7) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are [2005]

Smart Achievers · Accepted Answer

(4)

The given curve is $y = x^{2} + 6$

Equation of tangent at (1, 7) is $\frac{1}{2} (y + 7) = x \cdot 1 + 6 \Rightarrow 2 x - y + 5 = 0 ...(i)$

Since tangent (i) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ with centre $C (- 8, - 6)$ at Q.

$∴$ Equation of $C Q$ which is perpendicular to (i) is

$y + 6 = - \frac{1}{2} (x + 8) \Rightarrow x + 2 y + 20 = 0 ...(ii)$

On solving equations (i) and (ii), we get the coordinates of $Q$ as $x = - 6, y = - 7$

$∴$ Coordinates of $Q$ is $(- 6, - 7)$

Solution

Q.
Tangent to the curve $y = x^{2} + 6$ at a point (1, 7) touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are [2005]

1 $(- 6, - 11)$

2 $(- 9, - 13)$

3 $(- 10, - 15)$

4 $(- 6, - 7)$

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Solution

Q. Tangent to the curve y=x2+6 at a point (1, 7) touches the circle x2+y2+16x+12y+c=0 at a point Q. Then the coordinates of Q are [2005]

1 (-6,-11)

2 (-9,-13)

3 (-10,-15)