Suppose $a x + b y + c = 0$ , where $a, b, c$ are in AP be normal to a family of circles. The equation of the circle of the family intersecting the circle $x^{2} + y^{2} - 4 x - 4 y - 1 = 0$ orthogonally is

x^{2} + y^{2} - 2 x + 4 y - 3 = 0

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

x^{2} + y^{2} + 2 x - 4 y - 3 = 0

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

x^{2} + y^{2} - 2 x + 4 y - 5 = 0

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

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

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

Open in App

Solution

The correct option is A $x^{2} + y^{2} - 2 x + 4 y - 3 = 0$
Given, $a + c = 2 b \Rightarrow a - 2 b + c = 0$

Thus $a x + b y + c = 0$ will always pass through $(1, - 2)$

Normal to a circle will always passes through centre

$∴$ centre = $(1, - 2)$

Let the equation of family of circle

$x^{2} + y^{2} - 2 x + 4 y + c = 0$

This circle is orthogonal to $x^{2} + y^{2} - 4 x - 4 y - 1 = 0$

Condition of orthogonality

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

$2 (1) (- 2) + 2 (2) (- 2) = c_{1} - 1$

$4 - 8 = c_{1} + 1$

$c_{1} = - 3$

Equation of circle

$x^{2} + y^{2} - 2 x + 4 y - 3 = 0$

Suggest Corrections

Similar questions

The equation of the circle which intersects circles $x^{2} + y^{2} + x + 2 y + 3 = 0, x^{2} + y^{2} + 2 x + 4 y + 5 = 0$ and

$x^{2} + y^{2} - 7 x - 8 y - 9 = 0$ at right angle, will be

Q. The equation of the circle with centre at (1, -2) and passing through the centre of the given circle

x^{2} + y^{2} + 2 y - 3 = 0,

Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

Suppose

a x + b y + c = 0

, where a, b, c are in A. P. be normal to a family of circles. The equation of the circle of the family intersects the circle

x^{2} + y^{2} - 4 x - 4 y - 1 = 0

orthogonally is

Q. The circles

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

and

x^{2} + y^{2} + 4 x + 4 y - 1 = 0

Join BYJU'S Learning Program

Suppose ax+by+c=0, where a,b,c are in AP be normal to a family of circles. The equation of the circle of the family intersecting the circle x2+y2−4x−4y−1=0 orthogonally is

Suppose $a x + b y + c = 0$ , where $a, b, c$ are in AP be normal to a family of circles. The equation of the circle of the family intersecting the circle $x^{2} + y^{2} - 4 x - 4 y - 1 = 0$ orthogonally is