For the circle, #(x+1)^2+(y+2)^2=25# What is the length of the tangent to the circle from the point #(6,4)#?

1 Answer
Shwetank Mauria
Mar 25, 2018

The length of the tangent to the circle from the point #(6,4)# to circle #(x+1)^2+(y+2)^2=25# is #2sqrt15#

Explanation:

The circle #(x+1)^2+(y+2)^2=25# is a circle with center #(-1,-2)# and radius #5#.

As the length of the tangent from the external point, radius to the point at which tangent touches circle and line joining external point to center of circle form a right angle.

Now radius is #sqrt25=5#, distance between external point and center of circle is #sqrt((6+1)^2+(4+2)^2)#

and hence using Pythagorus theorem, lengt of tangent is #sqrt((6+1)^2+(4+2)^2-25)=sqrt(49+36-25)=sqrt60=2sqrt15#

graph{((x+1)^2+(y+2)^2-0.03)((x+1)^2+(y+2)^2-25)((x-6)^2+(y-4)^2-0.03)=0 [-10, 10, -4.8, 5.2]}

Obsserve that the length of the tangent to a circle #x^2+y^2+2gx+2fy+c=0# from an external point #(x_1,y_1)# will always be #sqrt(x_1^2+y_1^2+2gx_1+2fy_1+c)#.

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