Q 6 TB - Calculus

Question

To find the equation of circle in ${\mathrm{ℝ}}^2$having center at (-4, 3) and passing through (5, -2)

Step-by-Step Solution

Answer

The required equation is ${(x+4)}^2+{(y-3)}^2=106$

Given information

A circle in ${\mathrm{ℝ}}^2$ having center at (-4, 3) and passes through (5, -2).

Finding the equation

The equation as per the center given is:

${(x+4)}^2+{(y-3)}^2=r^2$

where $r$ is the radius

Putting the given point in the equation:

${(5+4)}^2+{(-2-3)}^2=r^2$

Simplifying:

$r^2=106$

The equation of circle is:

${(x+4)}^2+{(y-3)}^2=106$