The equation of the circle whose centre is 1-3 and which touches the line 2 x - y -4-0 isA- 5 x2-5 y2-10 x-30 y-49-0B- 5 x2-5 y2-10 x-30 y-49-0C- 5 x2-5 y2-10 x-30 y-49-0D- x2-y2-10 x-30 y-49-0

The correct option is A r = d = | 2+3 4/√(4 + 1)| = 1/√(5) (x 1)^2 + (y + 3)^2 = 1/5 ...

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Standard VIII

Mathematics

Tangent

The equation ...

Question

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

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x² + y² - 10x + 30y - 49 = 0

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Solution

The correct option is A

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

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

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

The correct option is A r = d = ∣∣∣2+3−4√4+1∣∣∣=1√5 (x−1)2+(y+3)2=15

The correct option is A

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

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