Question

Tangents are drawn from an external point `(x_1,y_1)` to the circle `x^2+y^2=a^2`. How do we find the angle between the two tangents?

Answer 1

`2*sin^-1(a/sqrt(x_1^2+y_1^2))`

Explanation:

Given equation of the circle : `x^2+y^2=a^2`,
`=> center =(0,0), radius =a`
Let `O` be the center of the circle, and `theta` be the angles between the two tangents `(angleAPB)`, as shown in the figure.
As `anglePAO=anglePBO=90^@, and AO=BO=a`
`=> DeltaPAO and DeltaPBO` are congruent.
`=> PA=PB, and PO` bisects `angleAPB`
`=> angleAPO=angleBPO=theta/2`

As `PO=sqrt(x_1^2+y_1^2)`
`=> sin(theta/2)=a/(PO)=a/sqrt(x_1^2+y_1^2)`

`=> theta/2=sin^-1(a/sqrt(x_1^2+y_1^2))`

`=> theta = 2* sin^-1(a/sqrt(x_1^2+y_1^2))`