Question

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)`?

Answer 1

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)`.