Find the equation to the circle which has its centre at the point $(1, - 3)$ and touches the straight line $2 x - y - 4 = 0$ .

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Solution

Give, the centre of the circle is $(1, - 3)$ and the equation of the line that it touches is $2 x - y - 4 = 0$
Therefore, radius of the circle is, $∣ ∣ ∣ ∣ ∣ \frac{2 \times 1 - (- 3) - 4}{\sqrt{(2^{2} + 1)}} ∣ ∣ ∣ ∣ ∣ = \frac{1}{\sqrt{5}}$

Therefore, the equation of the circle is
$(x - 1)^{2} + (y + 3)^{2} = {(\frac{1}{\sqrt{5}})}^{2} \Rightarrow 5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0$

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Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line $5 x + 12 y - 1 = 0$

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5 x + 12 y = 1

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

2 x + 3 y = 1,

2 x - y + 1 = 0

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x + y - 9 = 0

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