Chapter 12, Problem 5CR

To determine

The standard equation of the circle for the given information as center at point $\left(-1,2\right)$ and the point $\left(3,5\right)$ on the circle.

Answer to Problem 5CR

Solution:

The equation of circle with center $\left(-1,2\right)$ and the point $\left(3,5\right)$ on the circle is $x^2+y^2+2x-4y-20=0$.

Explanation of Solution

Given information:

Center of the circle is at $\left(-1,2\right)$ and the point on the circle is $\left(3,5\right)$.

Explanation:

From the given two points first find the radius,

$r=\sqrt{{\left(3-\left(-1\right)\right)}^2+{\left(5-2\right)}^2}$

$r=\sqrt{4^2+3^2}=\sqrt{16+9}$

$r=\sqrt{25}=5$

Thus, the radius of the circle is $5$.

Therefore the equation of the standard circle is given by,

${\left(x-\left(-1\right)\right)}^2+{\left(y-2\right)}^2=5^2$

$\Rightarrow {\left(x+1\right)}^2+{\left(y-2\right)}^2=5^2$

$\Rightarrow x^2+2x+1+y^2-4y+4=25$

$\Rightarrow x^2+y^2+2x-4y-20=0$

Therefore, the standard equation of the circle with center $\left(-1,2\right)$ and the point $\left(3,5\right)$ on the circle is $x^2+y^2+2x-4y-20=0$.