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777dan777 [17]
3 years ago
8

Find an equation of the circle that has center (-4,3) and passes through (6,-4).

Mathematics
1 answer:
kumpel [21]3 years ago
5 0

Answer:

(x+4)^2+(y-3)^2=149

Step-by-step explanation:

The equation of a circle is (x-h)^2+(y-k)^2=r^2 where (h,k) is the center of the circle and r is the radius of the circle.

Given that (h,k)\rightarrow(-4,3) and it passes (6,-4), their distance between each other must the radius of the circle, so we can use the distance formula to find the radius:

d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\d=\sqrt{(-4-3)^2+(6-(-4))^2}\\\\d=\sqrt{(-7)^2+10^2}\\\\d=\sqrt{49+100}\\\\d=\sqrt{149}

Therefore, if the length of the radius is r=\sqrt{149} units, then r^2=149, making the final equation of the circle (x+4)^2+(y-3)^2=149

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