Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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



Sagot :

Step-by-step explanation:

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

[tex](3 + 1) {}^{2} + ( - 2 + 6) {}^{2} = \sqrt{8} [/tex]

The diameter is sqr root of 8 units.

A circle equation is

[tex] {x}^{2} + {y}^{2} = {r}^{2} [/tex]

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

[tex]r = \frac{ \sqrt{8} }{2} = \frac{ \sqrt{8} }{ \sqrt{4} } = \sqrt{2} [/tex]

[tex] {r}^{2} = { \sqrt{2} }^{2} = 2[/tex]

So our radius is 2.

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

[tex] \frac{ - 6 - 2}{2} = - 4[/tex]

[tex] \frac{3 - 1}{2} = 1[/tex]

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

[tex](x + 4) {}^{2} + (y - 1) {}^{2} = ( \sqrt{2} ) {}^{2} [/tex]