Question

The point of intersection of the tangent $x+\sqrt{3}y=8$ with the circle $x^2+y^2=16$ is

• A. $(2\sqrt{2},2\sqrt{2})$
• B. $(2,2\sqrt{3})$
• C. $(2\sqrt{3},2)$
• D. $(2,-2\sqrt{3})$

Answer 1

The correct option is B $(2,2\sqrt{3})$

Point form of tangent to circle
$x{x}_1+y{y}_1=16\cdots \left(1\right)$
Tangent to the circle is $x+\sqrt{3}y=8$
Multiplying above equation by $2$, we get
$2x+2\sqrt{3}y=16\cdots \left(2\right)$
Comparing equation $\left(1\right)$ and $\left(2\right),$ we get
$x_1=2,y_1=2\sqrt{3}$