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3 years ago

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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.

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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