Question

Find an equation of the circle whose diameter has endpoints (-6, -1) and (-2,3).

Answer 1

Step-by-step explanation:

Let find the distance of the diameter. using distance formula.

$(3 + 1) {}^{2} + ( - 2 + 6) {}^{2} = \sqrt{8}$

The diameter is sqr root of 8 units.

A circle equation is

${x}^{2} + {y}^{2} = {r}^{2}$

where r is the radius. The radius is half the diameter so

$r = \frac{ \sqrt{8} }{2} = \frac{ \sqrt{8} }{ \sqrt{4} } = \sqrt{2}$

${r}^{2} = { \sqrt{2} }^{2} = 2$

So our radius is 2.

Now we need to find the midpoint or Center of the diameter.

$\frac{ - 6 - 2}{2} = - 4$

$\frac{3 - 1}{2} = 1$

So the center of the circle is (-4,1). So our equation of the Circle us

$(x + 4) {}^{2} + (y - 1) {}^{2} = ( \sqrt{2} ) {}^{2}$