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

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None of these

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Solution

The correct option is 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, $5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0.$

Suggest Corrections

Similar questions

(1, - 3)

2 x - y - 4 = 0

2 x - 3 y = 10, 2 x - 3 y = 4

2 x + 3 y = 1,

4 x - 3 y + 10 = 0

4 x - 3 y - 30 = 0

2 x + y = 0

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