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If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is ?

If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is ?

Grade:12

3 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
6 years ago
Hello student,
Let the centre be (α, β)
∵It cut the circle x2 + y2 = p2 orthogonally
2(-α) × 0 + 2(-β) × 0 = c1 – p2
c1 = p2
Let equation of circle is x2 + y2 - 2αx - 2βy + p2 = 0
It pass through (a, b) ⇒ a2+ b2- 2αa - 2βb + p2 = 0
Locus ∴ 2ax + 2by – (a2 + b2 + p2) = 0.
Thanks and Regards
Shaik Aasif
askIITians faculty
sajan Saran
17 Points
4 years ago
Let the centre be (α, β)It cut the circle x2 + y2 = p2 orthogonally2(-α) × 0 + 2(-β) × 0 = c – p2c = p2Let equation of circle is x2 + y2 + 2αx +2βy + p2 = 0It pass through (a, b) ⇒ a2+ b2+2αa +2βb + p2 = 0Locus ∴ 2ax + 2by + (a2 + b2 + p2) = 0.
Kushagra Madhukar
askIITians Faculty 629 Points
9 months ago
Dear student,
Please find the attached solution to your problem below.
 
Let the centre be (α, β)
∵It cut the circle x2 + y2 = p2 orthogonally
2(-α) × 0 + 2(-β) × 0 = c1 – p2
c1 = p2
Let equation of circle is x2 + y2 - 2αx - 2βy + p2 = 0
It passes through (a, b) ⇒ a2+ b2- 2αa - 2βb + p2 = 0
Locus ∴ 2ax + 2by – (a2 + b2 + p2) = 0.
 
Thanks and regards,
Kushagra

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