The locus of centre of circle passing through (a,b) and cuts orthogonally to circle $x^{2} + y^{2} = p^{2}$ is

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y x^{2} - 3 a x - 4 b y + (a^{2} + b^{2} + p^{2}) = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y x^{2} - 2 a x + 3 b y + (a^{2} - b^{2} + p^{2}) = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$
Let equation of circle be
Let equation of circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
with $x^{2} + y^{2} = p^{2}$ cutting orthogonoally
we get $0 + 0 = c - p^{2}$ and passes through (a,b)
we get $a^{2} + b^{2} + 2 g a + 2 f b + p^{2} = 0$
$\Rightarrow$ $2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$
Required locus as centre $(- g, - f) = (x, y)$

Suggest Corrections

Similar questions

(a, b)

x^{2} + y^{2} = k^{2}

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

The two circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch if $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

(a, b)

x^{2} + y^{2} = 4

m a x + m a y - (a^{2} + b^{2} + p^{2})

m

x^{2} + y^{2} + 2 a x - 2 b y + a^{2} - b^{2} = 0

x^{2} + y^{2} + 2 a x + c^{2} = 0

x^{2} + y^{2} + 2 b y + c^{2} = 0

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

Join BYJU'S Learning Program