Please answer this : - Economics - Unit I- Indian Economy 1950-1990

Question

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Rahul K · Accepted Answer

Dear Student,
Below provided solution is similar to the above question please go through the solution,

If the line 2x - y +1 = 0 touches the circle at the point (2,5) to the center is perpendicular, then

$y = 2 x + 1 S l o p e = \frac{- 1}{2}$

As, the line passes through (2,5) having $S l o p e = \frac{- 1}{2}$

$\Rightarrow y - 5 = \frac{- 1}{2} (x - 2) \Rightarrow 2 y - 10 = - x + 2 \Rightarrow 2 y = - x + 2 + 10 \Rightarrow 2 y = - x + 12 \Rightarrow y = - \frac{1}{2} x + 6........... (i)$
It is given that the center of the circle lies on the line x+y -9 =0........(ii)

So, this line passes through the center of the circle.

Put the value of y in (ii), we get,

$\Rightarrow x + (- \frac{1}{2} x + 6) = 9 \Rightarrow x + - \frac{1}{2} x + 6 = 9 \Rightarrow \frac{1}{2} x = 9 - 6 \Rightarrow \frac{1}{2} x = 3 \Rightarrow x = 6 F r o m (i), y = (- \frac{1}{2} \times 6) + 6 = - 3 + 6 \Rightarrow x = 3 ∴ (6, 3) i s t h e c e n t r e o f t h e c i r c l e .$
Now,

Distance from the centre of circle (6,3) to the point (2,5) which touches the circle.

$\sqrt{{(2 - 6)}^{2} + {(5 - 3)}^{2}} = \sqrt{{(- 4)}^{2} + {(2)}^{2}} = \sqrt{16 + 4} = \sqrt{20} T h u s, t h e e q u a t i o n o f a c i r c l e i s {(x - 6)}^{2} + {(y - 3)}^{2} = {(\sqrt{20})}^{2} \Rightarrow x^{2} + 36 - 12 x + y^{2} + 9 - 6 y = 20 \Rightarrow x^{2} + y^{2} - 12 x - 6 y + 45 = 20 \Rightarrow x^{2} + y^{2} - 12 x - 6 y + 25 = 0$
Regards.