Question

100 points need answer asap!! The diameter of Circle Q terminates on the circumference of the circle at (3,0) and (-4,0). Write the equation of the circle in standard form. Show all of your work for full credit.

Answer 1

Answer:

$4(x+0.5)^2+4y^2=49$

Step-by-step explanation:

Since the diameter of Circle Q terminates on the circumference of the circle at (3,0) and (-4,0), the diameter lies on the x-axis.

The midpoint of the ends of the diameter gives the center of the circle.

$(\frac{3+-4}{2},\frac{0+0}{2} =(\frac{-1}{2},\frac{0}{2})=(-0.5,0)$

The radius of this circle is the distance from the center to one of this points on the circumference.

$r=|3-0.5|=3.5$

The equation of the circle is given by $(x-a)^2+(x-b)^2=r^2$

where (a,b)=(-0.5,0) is the center and r=3.5

We substitute to get:

$(x--0.5)^2+(y-0)^2=3.5^2$

The standard form equation is $4(x+0.5)^2+4y^2=49$