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