Question

Write the equation of a circle with center (4, -5) and radius = square root of 13

Answer 1

Hello!

The answer is:

The equation of the given circle is:

$(x-4)^{2}+(y+5)^{2}=13$

Why?

The equation of a circle is given by the following equation:

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

We are given the center point (4,-5) and the radius.

So,

Where:

$h=x=4\\k=y=-5\\r=\sqrt{13}$

Then, substituting into the circle equation, we have:

$(x-4)^{2} +(y-(-5))^{2}=(\sqrt{13})^{2}$

$(x-4)^{2} +(y-(-5))^{2}=(\sqrt{13})^{2}$

Hence, the simplified equation of the circle is:

$(x-4)^{2}+(y+5)^{2}=13$

Have a nice day!

Answer 2

Answer:

The required equation in standard form is $(x-4)^2+(y+5)^2=13$

Step-by-step explanation:

The equation of a circle with center (h,k) an radius, r units is given by the formula;

$(x-h)^2+(y-k)^2=r^2$

The given circle has center (4,-5) and radius $r=\sqrt{13}$.

We substitute these values into the formula to obtain;

$(x-4)^2+(y--5)^2=(\sqrt{13})^2$

We simplify to get;

$(x-4)^2+(y+5)^2=13$