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

Sagot :

Answer:

[tex](x+4)^2+(y-3)^2=149[/tex]

Step-by-step explanation:

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

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

[tex]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}[/tex]

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

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