Answer:
B. x² + y² - 18x - 16y + 120 = 0
Step-by-step explanation:
You want the equation of the circle with radius 5 that is tangent to x² +y² -2x -4y -20 = 0 at the point (5, 5).
The standard form equation of a circle is ...
(x -h)² +(y -k)² = r² . . . . . . . . circle with center (h, k) and radius r
In general form, this is ...
x² +y² -2hx -2ky +(h² +k² -r²) = 0
The center (h, k) can be found from the coefficients of x and y.
Using the above information, we find the center of the given circle is ...
(h, k) = (-2/-2, -4/-2) = (1, 2)
This lies in the first quadrant. The tangent point (5, 5) is above and to the right of this, and is also in the first quadrant. That means the center of the desired circle will lie in the first quadrant, requiring both the coefficients of x and y in its equation to be negative. This eliminates choices C and D.
If we call the constant in the general form circle equation 'f', then we require ...
h² +k² -r² = f
For the equations of choices A and B, (h, k) = (9, 8), so the constant 'f' will be ...
9² +8² -5² = 81 +64 -25 = 120
The equation of the desired circle is x² + y² - 18x - 16y + 120 = 0, choice B.
