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ddd [48]
3 years ago
5

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.
Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0

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

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