The centre of a circle is (2, -3) and the cir | vedclass

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

Question

(C) Let the radius of the required circle be $r$. Given that the circumference is $10 \pi$. $2 \pi r = 10 \pi$ $\Rightarrow r = 5$ The centre of the c

Vedclass · Accepted Answer

$x^2 + y^2 - 4x + 6y - 12 = 0$

Vedclass · Answer

(C) Let the radius of the required circle be $r$. Given that the circumference is $10 \pi$. $2 \pi r = 10 \pi$ $\Rightarrow r = 5$ The centre of the circle is $(h, k) = (2, -3)$. The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. Substituting the values: $(x - 2)^2 + (y - (-3))^2 = 5^2$ $(x - 2)^2 + (y + 3)^2 = 25$ $x^2 - 4x + 4 + y^2 + 6y + 9 = 25$ $x^2 + y^2 - 4x + 6y + 13 - 25 = 0$ $x^2 + y^2 - 4x + 6y - 12 = 0$ Thus,option $C$ is correct.

The centre of a circle is $(2, -3)$ and the circumference is $10 \pi$. Then its equation is

Similar Questions

The equation of the smallest circle passing through the points $(2,2)$ and $(3,3)$ is

If the radius of a circle $x^{2}+y^{2}-4x+6y-k=0$ is $5$,then $k=$

If the centre of a circle is $(2, 3)$ and a tangent is $x + y = 1$,then the equation of this circle is

If two vertices of an equilateral triangle are $A(-a, 0)$ and $B(a, 0)$,where $a > 0$,and the third vertex $C$ lies above the $x$-axis,then the equation of the circumcircle of $\Delta ABC$ is:

If the vertices of a triangle are $(2, -2)$,$(-1, -1)$,and $(5, 2)$,then the equation of its circumcircle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module