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. Find the co-ordinates of another point $B$ on the circle such that $A B = 4$ .

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Solution

Let $A$ be $(h, k)$ tangent at which is $h x + k y = 8$ . It passes through $(4, 0)$ .
$∴ 4 h = 8$ or $h = 2$
Also $h^{2} + k^{2} = 8 ∴ k^{2} = 4$
$∴ k = 2$ (only) is Ist quadrant.
$∴ A (2, 2)$ . Now let $B$ be $(p, q)$
$∴ p^{2} + q^{2} = 8 . . . . . (1)$
Also $A B = 4 \Rightarrow (p - 2)^{2} + (q - 2)^{2} = 16$
or $p^{2} + q^{2} - 4 (p + q) + 8 = 16$
or $p + q = 0$ , by $(1) . . . . . (2)$
Solving $(1)$ and $(2)$ , we get $(2, - 2)$ or $(- 2, 2)$ .

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A tangent drawn from the point (4,0) to the circle x2+y2=8 touches it at a point A in the first quadrant. Find the co-ordinates of another point B on the circle such that AB=4.

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. Find the co-ordinates of another point $B$ on the circle such that $A B = 4$ .