Answer:
[tex](x-5)^2+(y+4)^2=100[/tex]
Step-by-step explanation:
The standard form of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex] .... (1)
where, (h,k) is center of the circle and r is the radius.
It is given that a circle is centered at the point (5, -4) and passes through the point (-3, 2).
[tex]h=5,k=-4[/tex]
Distance between (5, -4) and (-3, 2) is radius of the circle.
[tex]r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]r=\sqrt{(-3-5)^2+(2-(-4))^2}[/tex]
[tex]r=\sqrt{(-8)^2+(6)^2}[/tex]
On further simplification we get
[tex]r=\sqrt{64+36}[/tex]
[tex]r=\sqrt{100}[/tex]
[tex]r=10[/tex]
The radius of the circle is 10 units.
Substitute h=5,k=-4 and r=10 in equation (1).
[tex](x-5)^2+(y-(-4))^2=(10)^2[/tex]
[tex](x-5)^2+(y+4)^2=100[/tex]
Therefore, the equation of the circle is [tex](x-5)^2+(y+4)^2=100[/tex].
