Find the centre and radius of the circle x^{2 | vedclass

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

Question

(A) The given equation of the circle is $x^{2}+y^{2}-8x+10y-12=0$. Rearranging the terms,we get: $(x^{2}-8x) + (y^{2}+10y) = 12$. Completing the squar

Vedclass · Accepted Answer

$(4, -5), \sqrt{53}$

Vedclass · Answer

(A) The given equation of the circle is $x^{2}+y^{2}-8x+10y-12=0$. Rearranging the terms,we get: $(x^{2}-8x) + (y^{2}+10y) = 12$. Completing the square for $x$ and $y$: $(x^{2}-8x+16) + (y^{2}+10y+25) = 12+16+25$. This simplifies to: $(x-4)^{2} + (y+5)^{2} = 53$. Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,we identify the centre $(h, k) = (4, -5)$ and the radius $r = \sqrt{53}$.

Find the centre and radius of the circle $x^{2}+y^{2}-8x+10y-12=0$.

Similar Questions

The equation $ax^2 + 2y^2 + 2bxy + 2x - y + c = 0$ represents a circle passing through the origin,if:

If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$,then the equation of the circle is

$A$ point which lies on the circle passing through the points $(1,1), (-6,0),$ and $(-2,2)$ is

The equation of the circle passing through $(4, 5)$ and having the centre at $(2, 2)$ is:

If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8x-6y=0$,$4x^2+4y^2-4x-12y-186=0$,and $x^2+y^2-6x+6y-9=0$ respectively,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module