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Question: The locus of the centre of the circle which cuts off intercepts of length \(2 a\) and \(2 b\) from...

The locus of the centre of the circle which cuts off intercepts of length 2a2 a and 2b2 b from x-axis and y-axis respectively, is.

A

x+y=a+bx + y = a + b

B

x2+y2=a2+b2x ^ { 2 } + y ^ { 2 } = a ^ { 2 } + b ^ { 2 }

C

x2y2=a2b2x ^ { 2 } - y ^ { 2 } = a ^ { 2 } - b ^ { 2 }

D

x2+y2=a2b2x ^ { 2 } + y ^ { 2 } = a ^ { 2 } - b ^ { 2 }

Answer

x2y2=a2b2x ^ { 2 } - y ^ { 2 } = a ^ { 2 } - b ^ { 2 }

Explanation

Solution

2g2c=2a2 \sqrt { g ^ { 2 } - c } = 2 a ….(i)

2f2c=2b2 \sqrt { f ^ { 2 } - c } = 2 b ….(ii)

On squaring (i) and (ii) and then subtracting (ii) from (i), we get

Hence the locus is x2y2=a2b2x ^ { 2 } - y ^ { 2 } = a ^ { 2 } - b ^ { 2 }.