Chapter T, Problem 2BDT

To determine

To find: The equation of the circle that has center at $\left(-1,4\right)$ and passes through the point $\left(3,-2\right)$.

Answer to Problem 2BDT

The equation of the circle is ${\left(x+1\right)}^2+{\left(y-4\right)}^2=52$.

Explanation of Solution

Formula used:

The equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where the center of the circle is at the point $\left(h,k\right)$, r is the radius of the circle.

Calculation:

Suppose, the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

It is given that, the center of the circle is at the point $\left(-1,4\right)$.

Thus, substitute $h=-1$ and $k=4$ in ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ to obtain the equation of the straight line,

${\left(x+1\right)}^2+{\left(y-4\right)}^2=r^2$

It is given that, the line passes through the point $\left(2,-5\right)$.

Thus, the radius of the circle is the distance between the point $\left(2,-5\right)$ and the radius. That is,

$\begin{array}{c}{r}^2={\left(3-\left(-1\right)\right)}^2+{\left(-2-4\right)}^2\\ ={\left(4\right)}^2+{\left(-6\right)}^2\\ =16+36\\ =52\end{array}$

Thus, substitute $r^2=52$ in ${\left(x+1\right)}^2+{\left(y-4\right)}^2=r^2$ to obtain the equation of the straight line,

${\left(x+1\right)}^2+{\left(y-4\right)}^2=52$

Thus, the equation of the circle is ${\left(x+1\right)}^2+{\left(y-4\right)}^2=52$.