Question

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

• A. $x^2+y^2+4x+6y+12=0$
• B. $x^2+y^2-4x+6y+12=0$
• C. $x^2+y^2-4x+6y-12=0$
• D. $x^2+y^2-4x-6y-12=0$

Answer 1

Answer: (C)

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

The circumference of the circle is given as $10\pi$.

$C=2\pi r$

$10\pi =2\pi r$

$r=\frac{10\pi }{2\pi }$

$r=5$

The general form of the equation is,

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

The center of the circle is given as $\left(h,k\right)=\left(2,-3\right)$

${\left(x-2\right)}^2+{\left(y-\left(-3\right)\right)}^2={\left(5\right)}^2$

${\left(x-2\right)}^2+{\left(y+3\right)}^2=25$

$x^2+4-4x+y^2+9+6y=25$

$x^2+y^2-4x+6y+13=25$

$x^2+y^2-4x+6y-25+13=0$

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

Therefore, the equation of the circle is,

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