Find the centre and radius of each of the circle
(i) $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
(ii) $4 x^{2} + 4 y^{2} - 24 x - 8 y - 24 = 0$

Open in App

Solution

i) $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ .
$(x - 1)^{2} - 1 + (y + 2)^{2} - 4 - 4 = 0$
$(x - 1)^{2} + (y + 2)^{2} = 9 = 3^{2}$ .
radius $= 3$ , centre $(1, - 2)$

ii) $4 x^{2} + 4 y^{2} - 24 x - 8 y - 24 = 0$
$x^{2} + y^{2} - 6 x - 2 y - 6 = 0$
$(x - 3)^{2} - 9 + (y - 1)^{2} - 1 - 6 = 0$ .
$(x - 3)^{2} + (y - 1)^{2} = 16 = 4^{2}$
radius $= 4$ , centre $(3, 1)$

Suggest Corrections

Similar questions

Q. Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:
(i)

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

(ii)

x^{2} + 4 y^{2} - 4 x + 24 y + 31 = 0

(iii)

4 x^{2} + y^{2} - 8 x + 2 y + 1 = 0

(iv)

3 x^{2} + 4 y^{2} - 12 x - 8 y + 4 = 0

(v)

4 x^{2} + 16 y^{2} - 24 x - 32 y - 12 = 0

(vi)

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

Q. Find the centre and radius of the circle:

x^{2} + y^{2} - 6 x - 8 y + 24 = 0

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.

Q. Find the radical centre of the sets of 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.

Circles $x^{2} + y^{2} - 2 x - 4 y = 0 a n d x^{2} + y^{2} - 8 y - 4 = 0$

Join BYJU'S Learning Program

Find the centre and radius of each of the circle(i) x2+y2−2x+4y−4=0(ii) 4x2+4y2−24x−8y−24=0

i) x2+y2−2x+4y−4=0.(x−1)2−1+(y+2)2−4−4=0(x−1)2+(y+2)2=9=32.radius =3, centre (1,−2)ii) 4x2+4y2−24x−8y−24=0x2+y2−6x−2y−6=0(x−3)2−9+(y−1)2−1−6=0.(x−3)2+(y−1)2=16=42radius =4, centre (3,1)

Find the centre and radius of each of the circle
(i) $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
(ii) $4 x^{2} + 4 y^{2} - 24 x - 8 y - 24 = 0$