Question

The length of tangent from the point $(-1,2)$ to the circle $x^2+y^2-8x+5y-7=0$.

• A. $1$
• B. $4$
• C. $2$
• D. $6$

Answer 1

The correct option is B $4$

The given equation of the circle can be written as

$(x-4{)}^2+{(y+\frac{5}{2})}^2-16-\frac{25}{4}-7=0$

or, $(x-4{)}^2+{(y+\frac{5}{2})}^2=\frac{117}{4}$

So, centere of the circle is $(4,-\frac{5}{2})$ and radius is $\frac{\sqrt{117}}{2}$ units.

Now distance between the point $(-1,2)$ and $(4,-\frac{5}{2})$ is $\frac{181}{2}$ units which is greater than the radius of the circle.

So, the point $(-1,2)$ lies outside of the circle.

The length of the tangent from the point $(-1,2)$ to the circle $x^2+y^2-8x+5y-7=0$ is $\sqrt{(-1{)}^2+2^2+8+10-7}=\sqrt{16}=4$ units.