SATHEE: Circle 3 Question 5

Circle 3 Question 5

5. If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^{2} + y^{2} + 4 x - 6 y = 12$ externally at the point $(1, - 1)$ , then the radius of $C$ is

(2019 Main,10 Jan I)

(a) 5

(b) $2 \sqrt{5}$

(d) 4

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Answer:

Correct Answer: 5. (a)

Solution:

Equation of tangent to the circle $x^{2} + y^{2} + 4 x - 6 y - 12 = 0$ at $(1, - 1)$ is given by $x x_{1} + y y_{1} + 2 (x + x_{1}) - 3 (y + y_{1}) - 12 = 0$ , where $x_{1} = 1$ and $y_{1} = - 1$

$\Rightarrow x - y + 2 (x + 1) - 3 (y - 1) - 12 = 0$

$\Rightarrow 3 x - 4 y - 7 = 0$

This will also a tangent to the required circle.

Now, equation of family of circles touching the line $3 x - 4 y - 7 = 0$ at point $(1, - 1)$ is given by

$(x - 1)^{2} + (y + 1)^{2} + λ (3 x - 4 y - 7) = 0$

So, the equation of required circle will be $(x - 1)^{2} + (y + 1)^{2} + λ (3 x - 4 y - 7) = 0$ , for some $λ \in R$ …(i)

$∵$ The required circle passes through $(4, 0)$

$∴ (4 - 1)^{2} + (0 + 1)^{2} + λ (3 \times 4 - 4 \times 0 - 7) = 0$

$\Rightarrow 9 + 1 + λ (5) = 0 \Rightarrow λ = - 2$

Substituting $λ = - 2$ in Eq. (i), we get

$(x - 1)^{2} + (y + 1)^{2} - 2 (3 x - 4 y - 7) = 0$

$\Rightarrow x^{2} + y^{2} - 8 x + 10 y + 16 = 0$

On comparing it with

$\begin{aligned} x^{2} + y^{2} + 2 g x + 2 f y + c = 0, we get \\ g = - 4, f = 5, c = 16 \\ ∴ & Radius = \sqrt{g^{2} + f^{2} - c} = \sqrt{16 + 25 - 16} = 5 \end{aligned}$

Circle 3 Question 5

5. If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^{2} + y^{2} + 4 x - 6 y = 12$ externally at the point $(1, - 1)$ , then the radius of $C$ is

Answer:

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Circle 3 Question 5

5. If a circle C passing through the point (4,0) touches the circle x2+y2+4x−6y=12 externally at the point (1,−1), then the radius of C is

Answer:

Engineering Preparation

Medical Preparation

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Other Resources

Syllabus

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Sample Mock Test

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NCERT Exercise Video Solution

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5. If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^{2} + y^{2} + 4 x - 6 y = 12$ externally at the point $(1, - 1)$ , then the radius of $C$ is