Find the centre and the radius of the circle | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation is $x^{2}+y^{2}+8x+10y-8=0$.Rearranging the terms,we get $(x^{2}+8x) + (y^{2}+10y) = 8$.Completing the squares,we add $(8/2)^{2

Vedclass · Accepted Answer

Centre: $(-4, -5)$,Radius: $7$

Vedclass · Answer

(A) The given equation is $x^{2}+y^{2}+8x+10y-8=0$.Rearranging the terms,we get $(x^{2}+8x) + (y^{2}+10y) = 8$.Completing the squares,we add $(8/2)^{2} = 16$ and $(10/2)^{2} = 25$ to both sides:$(x^{2}+8x+16) + (y^{2}+10y+25) = 8+16+25$.This simplifies to $(x+4)^{2} + (y+5)^{2} = 49$.Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,where $(h, k)$ is the centre and $r$ is the radius:$h = -4$,$k = -5$,and $r^{2} = 49$,so $r = 7$.Thus,the centre is $(-4, -5)$ and the radius is $7$.

Find the centre and the radius of the circle $x^{2}+y^{2}+8x+10y-8=0$.

Similar Questions

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^2 + y^2 = r^2$ at $A$ and $B$. Then $PA \cdot PB$ is equal to

Find the equation of the circle with centre $(0,2)$ and radius $2$.

The equation $2x^2 + 2y^2 + 4x + 8y + 15 = 0$ represents

The radius of the circle $2 x^2 + 2 y^2 - 3 x + 2 y - 1 = 0$ is $.......$ units.

If the equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle passing through the origin,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module