Find the radius of the circle whose equation is $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ .

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

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

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

Open in App

Solution

The correct option is B 3
Given,
Equation of the circle = $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$
$x^{2} - 2 x + y^{2} - 4 y + (1 + 4 - 9) = 0$
$(x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) - 9 = 0$
$(x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) = 9$
$(x - 1)^{2} + (y - 2)^{2} = 3^{2}$
Comparing the above equation with the general form of circle, $(x - h)^{2} + (y - k)^{2} = r^{2}$ , we get,
r = radius = 3 units.

Suggest Corrections

Similar questions

Q. Find radius of he circle whose equation is

x^{2} + y^{2} + 8 x + 4 y + 8 = 0

Q. Find the center and radius of the circle whose equation is given by

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

Q. Find the equation of the circle passing through the points of intersection of the circles

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

and

x^{2} + y^{2} - 10 x - 12 y + 40 = 0

and whose radius is 4

The radius of the circle S is same as the radius of $x^{2} + y^{2} - 2 x + 4 y - 11 = 0$ and the centre of S is the centre of $x^{2} + y^{2} - 2 x - 4 y + 11 = 0$ . Find the equation of S.

Find the equation of family of circles passing through the point of intersection of the circles $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ and $x^{2} + y^{2} - 10 x - 12 y + 40 = 0$ and whose radius is 4.

Join BYJU'S Learning Program

Find the radius of the circle whose equation is x2 + y2 − 2x − 4y − 4 = 0.

Find the radius of the circle whose equation is $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ .