Question

The endpoints of a diameter of a circle are (5,-10) and (-7,2). What is the equation of the circle in standard form?

Answer 1

Answer:

$(x+1)^2+(y+4)^2=(6\sqrt{2})^2$

Step-by-step explanation:

The length of the diameter$(d)$ will be the distance between $(5,-10)\ and\ (-7,2).$

$d=\sqrt{(2-(-10))^2+(-7-5)^2}=\sqrt{(12)^2+(-12)^2}=\sqrt{144+144}=\sqrt{288}\\\\d=12\sqrt{2}$

Radius:

$radius(r)=\frac{diameter}{2}=\frac{12\sqrt{2}}{2}=6\sqrt{2}$

Centre:

Let $(a,b)$ be centre of the circle.

Centre will be the mid point of end points of diameter.

$a=\frac{5-7}{2}=\frac{-2}{2}=-1\\\\b=\frac{-10+2}{2}=\frac{-8}{2}=-4\\\\Centre=(-1,-4)$

Equation of circle:

If $(a,b)$ be centre and $r$ be the radius.

Equation of circle: $(x-a)^2+(y-b)^2=r^2$

Here $(a,b)=(-1,-4)\ and\ r=6\sqrt{2}$

$(x-(-1))^2+(y-(4))^2=(6\sqrt{2})^2\\(x+1)^2+(y+4)^2=(6\sqrt{2})^2$