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 on another point $B$ on the circle such that $AB=4$, are

• A. $(2,-2)$
• B. $(-2,2)$
• C. $(2,2)$
• D. $(-2,-2)$

Answer 1

Answer: (A), (B)

$(-2,2)$
$(2,-2)$

Equation of given circle is $x^2+y^2=8$ ...(1)

Let $A(h,k)$ be the point of contact, in the first quadrant, of tangent from $P(4,0)$ to the circle (1).

Equation of tangent at $A(h,k)$ is $hx+ky=8.$

It passes through $P(4,0)$,

$\therefore 4h=8\Rightarrow h=2$

Since, $A(h,k)$ lies on the circle, we get

$h^2+k^2=8\Rightarrow 4+k^2=0\Rightarrow k=2(\therefore k>0)\Rightarrow A\equiv (2,2)$

Let the coordinate of point B on circle (1), be $(a,b)$ such that $AB=4$.

$\therefore a^2+b^2=8$ ...(2)

and ${AB}^2={(a-2)}^2+{(b-2)}^2=16$ ...(3)

Solving (2) and (3), we get $a=2,b=-2$ or $a=-2,b=2$

Hence the coordinates of $B$ are $(2,-2)$ or $(-2,2)$