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Standard VI

Mathematics

Collinear Points

The point on ...

Question

The point on the straight line x + y = 2, which is nearest to the circle $x^{2} + y^{2} - 10 x + 2 y + 22 = 0$

(1, 1)

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(4, -2)

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(-2, 9)

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(3, -1)

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Solution

The correct option is B (4, -2)
Shortest distance between $2$ curves occurs along the common normal. Let required point be $(t, 2 - t)$ . Normal at this will be $y (2 - t) = x - t .$ Since it passes through centre we get $- 1 - 2 + t = 5 - t$ $\Rightarrow$ t $= 4.$
So pt $= (4, - 2)$

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The point on the straight line x + y = 2, which is nearest to the circle x2+y2−10x+2y+22=0

The correct option is B (4, -2)Shortest distance between 2 curves occurs along the common normal. Let required point be (t,2−t). Normal at this will be y(2−t)=x−t. Since it passes through centre we get −1−2+t=5−t ⇒t =4.So pt =(4,−2)

The point on the straight line x + y = 2, which is nearest to the circle $x^{2} + y^{2} - 10 x + 2 y + 22 = 0$

The correct option is B (4, -2)
Shortest distance between $2$ curves occurs along the common normal. Let required point be $(t, 2 - t)$ . Normal at this will be $y (2 - t) = x - t .$ Since it passes through centre we get $- 1 - 2 + t = 5 - t$ $\Rightarrow$ t $= 4.$
So pt $= (4, - 2)$