SATHEE: Circle 2 Question 9

Circle 2 Question 9

10. If the circle $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is

(a) 2 or $- 3 / 2$

(b) -2 or $- 3 / 2$

(d) -2 or $3 / 2$

$(2000, 2 M)$

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Answer:

Correct Answer: 10. (a)

Solution:

Since, the given circles intersect orthogonally.

$\begin{aligned} ∴ & 2 (1) (0) + 2 (k) & (k) = 6 + k \\ [∵ 2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}] \\ 2 k^{2} - k - 6 = 0 \Rightarrow & k = - \frac{3}{2}, 2 \end{aligned}$

Circle 2 Question 9

10. If the circle $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is

Answer:

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Circle 2 Question 9

10. If the circle x2+y2+2x+2ky+6=0 and x2+y2+2ky+k=0 intersect orthogonally, then k 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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10. If the circle $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is