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Position of a Point with Respect to Circle

If x and y re...

Question

If x and y (real) satisfy the relation $x^{2} + y^{2} = 6 x - 8 y$ then let maximum of y be"k" and minimum of y be "n".similarly maximum of x be "p" and minimum of x be "h".The value of $k + n + p + h$ = $- 2$ .( Enter 1 if true or 0 otherwise)

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Solution

We have,
$x^{2} + y^{2} = 6 x - 8 y$
$\Rightarrow (x^{2} - 6 x + 9) + (y^{2} + 8 y + 16) = 25$
$\Rightarrow (x - 3)^{2} + (y + 4)^{2} = 5^{2}$
Clearly this is a equation of circle centered at $(3, - 4)$ with radius $5$
So any point on this circle can be taken as,
$x - 3 = 5 cos θ, y + 4 = 5 sin θ$
$\Rightarrow x = 3 + 5 cos θ, y = - 4 + 5 sin θ$
Ans since $- 1 \leq sin θ, cos θ \leq 1$
Hence $y_{m a x} = 3 + 5 (1) = 8, y_{m i n} = 3 + 5 (- 1) = - 2$
and $x_{m a x} = - 4 + 5 (1) = 1, x_{m i n} = - 4 + 5 (- 1) = - 9$
Therefore $k + n + p + h = - 2$

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