Question

A tangent drawn from the point (4, 0) to the circle $x^2+y^2=8$ touches it at a point A in the first quadrant. The coordinates of another point B on the circle such that $AB$ = 4 are

• A. $(2,-2)$
• B. $(-2,2)$
• C. $(-2\sqrt{2},0)$
• D. $(0,-2\sqrt{2})$

Answer 1

Answer: (A)

$(2,-2)$

$x^2+y^2=(2\sqrt{2}{)}^2,C=(0,0),r=2\sqrt{2}$

Let $y=mx+2\sqrt{2}\sqrt{m^2+1}$ be a tangent

To find the tangent through $(4,0)$ substitute into the equation

$⟹0=4m+2\sqrt{2}\sqrt{m^2+1}$

$⟹16m^2=8(m^2+1)$

$⟹m=-1$

Equation is

$y=-x+4$

Substituting in circle equation

$x^2+(-x+4{)}^2=8$

$⟹2x^2–8x+8=0$

$⟹x=2⟹y=2$

$A=(2,2)$

Any point on the circle is given be$(2\sqrt{2}\cos \theta ,2\sqrt{2}\sin \theta )$

Let B = $(2\sqrt{2}\cos {\theta }_1,2\sqrt{2}\sin {\theta }_1)$

$AB=4⟹A{B}^2=16$

$⟹(2\sqrt{2}\cos {\theta }_1,-2{)}^2+(2\sqrt{2}\sin {\theta }_1-2{)}^2=16$

$⟹8+4+4–4\sqrt{2}(\cos {\theta }_1+\sin {\theta }_1)=16$

$⟹\sin {\theta }_1+\cos {\theta }_1=0$

${\theta }_1=\frac{-\pi }{4}$

$B=(2\sqrt{2}\times \frac{1}{\sqrt{2}}2\sqrt{2}\times \frac{-1}{\sqrt{2}})=(2,-2)$