Find the points on the circle `x^2+y^2-2x+4y-20=0` which are the farthest and nearest to the point `(-5,6)dot`

Explanation

The given circle is
`x^(2)+y^(2)-2x+4y-20=0`
The center of the circle is C(1,-2) and the radius is `r=5`.
Here, point P(-5,6) lies outside the circle. Now,

Slope of CP `=(6-(-2))/(-5-1)=-(4)/(3)=tan theta`(say)
Points A and B lie at distance of 5 units from the center C.
Now, points a distance 5 units from C on the line CP are
`(1+-5 cos theta,-2+-5 sin theta)`
or `(1+-5(-(3)/(5)),-2+-5((4)/(5)))`
or `(1overset(-)(+)3,-2+-4)`
or `(-2,2),(4,-6)`
Clearly, point A(-2,2) is nearest to P and B (4,-6) is farthest from P.
Alternative method `:`
`CP=10`
Now, point A divides CP in ratio
`(AP)/(AC)=(CP-AC)/(AC)=(10-5)/(5)=1`
Therefore, the coordinates of A are
`((-5+1)/(2),(6-2)/(2))=(-2,2)`
Also, C is the midpoint of AB. Hence, the coordinates of B are `(4,-6)`