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Find the shortest distance from (10,7) to the circle x^2 y^2-4x-2y-20

Sagot :

Answer:

5,15

Explanation:

The equation of the given circle is x2+y2−4x−2y−20=0 and when substituted x=10 & y=7, its value becomes 100+49−40−14−20=75 which is greater than zero.

Thus, the point (10,7) lies outside the circle.

Its distance from the centre of the circle (2,1) is (10−2)2+(7−1)2=100=10 unit.

The minimum distance from the circle therefore becomes 10−5(i.e. radius) =5 and the maximum distance becomes 10+5=15 unit.