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Crank
3 years ago
12

Write the equation of a circle with center (6,4) that passes through the coordinate (2,1). In your final answer, include all of

your calculations.
Mathematics
2 answers:
kari74 [83]3 years ago
8 0
R^2=(x-6)^2+(y-4)^2

r^2=(6-2)^2+(4-1)^2, r^2=16+9=25

(x-6)^2+(y-4)^2=25
Phantasy [73]3 years ago
6 0

Answer:

(x-6)^2+(y-4)^2=25

Step-by-step explanation:

It is given that,

Center of the circle  = (6,4)

Circle passes through the point = (2,1)

Radius of the circle is distance between center (6,4) and point on he circle (2,1).

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

r=\sqrt{(2-6)^2+(1-4)^2}

r=\sqrt{(-4)^2+(-3)^2}

r=\sqrt{16+9}

r=\sqrt{25}

r=5

Standard form of a circle is

(x-h)^2+(y-k)^2=r^2

where, r is radius and (h,k) is center.

Substitute h=6, k=4 and r=5 in the above equation.

(x-6)^2+(y-4)^2=5^2

(x-6)^2+(y-4)^2=25

Therefore, the equation of circle is (x-6)^2+(y-4)^2=25.

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