The equation of the circle whose centre is (1, -3) and which touches the line

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Question

The equation of the circle whose centre is (1, -3) and which touches the line 2x - y - 4 = 0 is

Options:

A .

5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0

B .

5 x^{2} + 5 y^{2} + 10 x - 30 y + 49 = 0

C .

5 x^{2} + 5 y^{2} - 10 x + 30 y - 49 = 0

D . None of these

Answer: Option A
:
A
Radius of circle the required circle =

∣ ∣ ∣ \frac{2 + 3 - 4}{\sqrt{5}} ∣ ∣ ∣ = \frac{1}{\sqrt{5}}

Therefore the equation is,

(x - 1)^{2} + (y + 3)^{2} = \frac{1}{5}

i.e,

x^{2} + y^{2} - 2 x + 6 y + 1 + 9 = \frac{1}{5}

i.e,

5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0.

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