SATHEE: Circle 3 Question 6

Circle 3 Question 6

6. If the tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ , then the value of $c$ is

(a) 195

(b) 185

(d) 95

(2018 Main)

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Answer:

Correct Answer: 6. (d)

Solution:

Key Idea Equation of tangent to the curve

$x^{2} = 4 a y$ at $(x_{1}, y_{1})$ is $x x_{1} = 4 a \frac{y + y_{1}}{2}$

Tangent to the curve $x^{2} = y - 6$ at $(1, 7)$ is

$x = \frac{y + 7}{2} - 6$

$\Rightarrow 2 x - y + 5 = 0$

Equation of circle is $x^{2} + y^{2} + 16 x + 12 y + c = 0$

Centre $(- 8, - 6)$

$r = \sqrt{8^{2} + 6^{2} - c} = \sqrt{100 - c}$

Since, line $2 x - y + 5 = 0$ also touches the circle.

$\begin{array}{lc} ∴ & \sqrt{100 - c} = \frac{2 (- 8) - (- 6) + 5}{\sqrt{2^{2} + 1^{2}}} \\ \Rightarrow & \sqrt{100 - c} = \frac{- 16 + 6 + 5}{\sqrt{5}} \\ \Rightarrow & \sqrt{100 - c} = | - \sqrt{5} | \\ \Rightarrow & 100 - c = 5 \\ \Rightarrow & c = 95 \end{array}$

Circle 3 Question 6

6. If the tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ , then the value of $c$ is

Answer:

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Circle 3 Question 6

6. If the tangent at (1,7) to the curve x2=y−6 touches the circle x2+y2+16x+12y+c=0, then the value of c is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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6. If the tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ , then the value of $c$ is