To find: The equation of the circle that has center at (−1,4) and passes through the point (3,−2).

Answer to Problem 2BDT
The equation of the circle is (x+1)2+(y−4)2=52.
Explanation of Solution
Formula used:
The equation of the circle is (x−h)2+(y−k)2=r2, where the center of the circle is at the point (h,k), r is the radius of the circle.
Calculation:
Suppose, the equation of the circle is (x−h)2+(y−k)2=r2.
It is given that, the center of the circle is at the point (−1,4).
Thus, substitute h=−1 and k=4 in (x−h)2+(y−k)2=r2 to obtain the equation of the straight line,
(x+1)2+(y−4)2=r2
It is given that, the line passes through the point (2,−5).
Thus, the radius of the circle is the distance between the point (2,−5) and the radius. That is,
r2=(3−(−1))2+(−2−4)2=(4)2+(−6)2=16+36=52
Thus, substitute r2=52 in (x+1)2+(y−4)2=r2 to obtain the equation of the straight line,
(x+1)2+(y−4)2=52
Thus, the equation of the circle is (x+1)2+(y−4)2=52.
Chapter T Solutions
Bundle: Single Variable Calculus: Concepts And Contexts, 4th + Student Solutions Manual
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