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

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^{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!

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

x^{2} + y^{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 B $2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$
Let equation of the circle be,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0. (1)$
It passes through $(a, b)$
$∴ a^{2} + b^{2} + 2 g a + 2 f b + c = 0 (2)$
Since
$(1)$ cuts the circle $x^{2} + y^{2} = p^{2}$ orthogonally,
we have
$2 g x \times 0 + 2 f \times 0 = c - p^{2} \Rightarrow c = p^{2}$
so from
$(2)$ . we get $a^{2} + b^{2} + 2 g a + 2 f b + p^{2} = 0,$
Hence locus
of the centre $(- g, - f)$ of $(1)$ is $2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$

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

(a, b)

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

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

m

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

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