If the power of point $(1, - 2)$ with respect to $x^{2} + y^{2} = 1$ is equal to the radius of a circle and $(3, 2)$ is the centre of that circle, then the equation of that circle is

x^{2} + y^{2} - 6 x - 4 y - 3 = 0

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

x^{2} + y^{2} - 6 x - 4 y - 5 = 0

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

x^{2} + y^{2} - 6 x - 4 y + 3 = 0

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

x^{2} + y^{2} - 6 x - 4 y + 5 = 0

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

Open in App

Solution

The correct option is A $x^{2} + y^{2} - 6 x - 4 y - 3 = 0$
The power of the point $(1, - 2)$ with respect to the circle $x^{2} + y^{2} = 1$ is
$S_{1} = (1)^{2} + (- 2)^{2} - 1 = 4$
$∴$ radius, $r = 4$
Then the equation of circle whose centre $C \equiv (3, 2)$ and radius $r = 4$ is
$(x - 3)^{2} + (y - 2)^{2} = 4^{2}$
$\Rightarrow x^{2} + y^{2} - 6 x - 4 y - 3 = 0$

Suggest Corrections

Similar questions

Q. Equation of the circle passing through the point

(1, 1)

and point of intersection of circles

x^{2} + y^{2} = 6

and

x^{2} + y^{2} - 6 x + 8 = 0

Find the equation of director circle of the circle whose diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area is 154 square units.

Q. Consider three circles whose equations are

x^{2} + y^{2} + 3 x + 2 y + 1 = 0

x^{2} + y^{2} - x + 6 y + 5 = 0

and

x^{2} + y^{2} + 5 x - 8 y + 15 = 0

, then

Q. Tangents drawn from the point P(1, 8) to the circle

x^{2} + y^{2} - 6 x - 4 y - 11 = 0

touch the circle at points A and B. The equation of the circumcircle of triangle PAB is

Tangents are drawn from the point P(1, 8) to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B, then equation of the circumcircle of the triangle PAB is

Join BYJU'S Learning Program

If the power of point (1,−2) with respect to x2+y2=1 is equal to the radius of a circle and (3,2) is the centre of that circle, then the equation of that circle is

The correct option is A x2+y2−6x−4y−3=0The power of the point (1,−2) with respect to the circle x2+y2=1 is S1=(1)2+(−2)2−1=4 ∴ radius, r=4 Then the equation of circle whose centre C≡(3,2) and radius r=4 is (x−3)2+(y−2)2=42 ⇒x2+y2−6x−4y−3=0

If the power of point $(1, - 2)$ with respect to $x^{2} + y^{2} = 1$ is equal to the radius of a circle and $(3, 2)$ is the centre of that circle, then the equation of that circle is