The difference of the maximum and the minimum distance of a point on the line $x + y - 1 = 0$ , where it cuts x-axis, with respect to the circle $x^{2} + y^{2} - 6 x - 8 y + 24 = 0$ , is?

1

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2

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2 \sqrt{5}

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3 \sqrt{5}

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Solution

The correct option is B $1$
$x + y - 1 = 0$ cuts $x -$ axis at $(1, 0)$
Given circle : $x^{2} + y^{2} - 6 x - 8 y + 24 = 0$
$∴$ centre $= (- 3, - 4)$
radius $= \sqrt{25 - 24} = 1$
Distance of centre from point $= \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2}$
$∴$ Min distance $= 4 \sqrt{2} - 1$
Max distance $= 4 \sqrt{2} + 1$
Diff $= 2$ [B]

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Similar questions

What is the maximum and minimum distance of the point (9, 12) from the circle $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$

x^{2} + y^{2} + 6 x - 1 = 0

x^{2} + y^{2} - 3 y + 2 = 0

x^{2} + y^{2} + x + y - 3 = 0

P (1, 2)

θ

x -

x^{2} + y^{2} - 6 x - 8 y + 24 = 0

A

B

P A + P B = 4 \sqrt{2} .

cos θ + sin θ

(1, - 2)

x^{2} + y^{2} = 1

(3, 2)

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