SATHEE: Circle 2 Question 14

Circle 2 Question 14

15. If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = k^{2}$ orthogonally, then the equation of the locus of its centre is

$(1988, 2 M)$

(a) $2 a x + 2 b y - (a^{2} + b^{2} + k^{2}) = 0$

(b) $2 a x + 2 b y - (a^{2} - b^{2} + k^{2}) = 0$

(d) $x^{2} + y^{2} - 2 a x - 3 b y + (a^{2} - b^{2} - k^{2}) = 0$

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Answer:

Correct Answer: 15. (a)

Solution:

Let $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , cuts $x^{2} + y^{2} = k^{2}$ orthogonally.

$\begin{aligned} \Rightarrow & 2 g_{1} g_{2} + 2 f_{1} f_{2} & = c_{1} + c_{2} \\ \Rightarrow & - 2 g \cdot 0 - 2 f \cdot 0 & = c - k^{2} \\ \Rightarrow & c & = k^{2} \end{aligned}$

Also, $x^{2} + y^{2} + 2 g x + 2 f y + k^{2} = 0$ passes through $(a, b)$ .

$∴ a^{2} + b^{2} + 2 g a + 2 f b + k^{2} = 0$

$\Rightarrow$ Required equation of locus of centre is

$\begin{aligned} - 2 a x - 2 b y + a^{2} + b^{2} + k^{2} & = 0 \\ 2 a x + 2 b y - (a^{2} + b^{2} + k^{2}) & = 0 \end{aligned}$

Circle 2 Question 14

15. If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = k^{2}$ orthogonally, then the equation of the locus of its centre is

Answer:

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Circle 2 Question 14

15. If a circle passes through the point (a,b) and cuts the circle x2+y2=k2 orthogonally, then the equation of the locus of its centre 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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15. If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = k^{2}$ orthogonally, then the equation of the locus of its centre is