Example 13 Find the equation of the circle that is tangent to the circle C:x2+y2−6x−8y+17=0 at the point (1,2) and has a radius of 252.
Official solution
Consider the point (1,2) as the point circle curve Γ0:(x−1)2+(y−2)2=0. By property 9, the equation of the required circle is x2+y2−6x−8y−17+λ[(x−1)2+(y−2)2]=0. That is, (x−1+λ3λ)2+(y−1+λ4+2λ)2=(1+λ)28. From (1+λ)28=(252)2, we get λ1=−51 or λ2=−59. Thus, we obtain C1:(x−27)2+(y−29)2=(252)2 and C2:(x+23)2+(y+21)2=(252)2. It is easy to see that the point (1,2) satisfies the equation of circle C1, and the distance between the centers of circle C1 and circle C equals the absolute value of the difference of their radii, so circle C1 is internally tangent to circle C at point (1,2), and circle C1 is the required one. Similarly, circle C1 is externally tangent to circle C at point (1,2), and circle C2 is also the required one.
Source: NuminaMath-1.5,
licensed Apache-2.0.
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