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.

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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

Answer:

[tex](x-6)^2+(y-4)^2=25[/tex]

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).

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

[tex]r=\sqrt{(2-6)^2+(1-4)^2}[/tex]

[tex]r=\sqrt{(-4)^2+(-3)^2}[/tex]

[tex]r=\sqrt{16+9}[/tex]

[tex]r=\sqrt{25}[/tex]

[tex]r=5[/tex]

Standard form of a circle is

[tex](x-h)^2+(y-k)^2=r^2[/tex]

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

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

[tex](x-6)^2+(y-4)^2=5^2[/tex]

[tex](x-6)^2+(y-4)^2=25[/tex]

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