Question

Given that a circle has an equation of x^2+y^2+2x-6y-6=0, what is the exact length of AB given that a tangent is drawn from A(6,5) and touches the circle at B?

Answer 1

`abs(AB)=sqrt(37)`

Explanation:

Rewriting `x^2+y^2+2x-6y-6=0`
into standard form for a circle:

`color(white)("XXX")x^2+2xcolor(orange)(+1)=y^2-6ycolor(orange)(+9) = 6 color(orange)(+1+9)`

`color(white)("XXX")(x+1)^2+(y-3)^2=4^2`
which is the equation for a circle with center `C` at `(-1,3)`
and radius `r=4`

Remember that the tangent to a circle forms a right angle with the radius of the circle to the tangent point.

`abs(AC)=sqrt((6-(-1))^2+(5-3)^2)=sqrt(49+4)=sqrt(53)`

`abs(BC)= "the radius" = 4`

`abs(AB)^2= abs(AC)^2-abs(BC)^2`

`color(white)("XXX")=53-16 = 37`

`abs(AB)=sqrt(37)`