Equation of the circle- centred at 1-5 and touching the line 3 x -4 y -8- is A- x2-y2-2 x-10 y-1-0B- x2-y2-4 x-5 y-25-0C- x 2- y 2-2 x -10 y -26-0D- x 2- y 2-10 x -2 y -1-0

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Position of a Line with Respect to Circle

Equation of t...

Question

Equation of the circle, centred at $(1, - 5)$ and touching the line $3 x + 4 y = 8,$ is

x^{2} + y^{2} - 2 x + 10 y + 1 = 0

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x^{2} + y^{2} - 4 x + 5 y + 25 = 0

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x^{2} + y^{2} + 2 x - 10 y + 26 = 0

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x^{2} + y^{2} - 10 x + 2 y + 1 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 2 x + 10 y + 1 = 0$
Let $r$ be the radius of required circle.
Then $r$ is perpendicular distance of point $(1, - 5)$ from $3 x + 4 y - 8 = 0$
$\Rightarrow r = \frac{| 3 (1) + 4 (- 5) - 8 |}{\sqrt{3^{2} + 4^{2}}} \Rightarrow r = 5$
Hence required equation of circle is,
$(x - 1)^{2} + (y + 5)^{2} = 5^{2} \Rightarrow x^{2} + y^{2} - 2 x + 10 y + 1 = 0$

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Equation of the circle, centred at (1,–5) and touching the line 3x+4y=8, is

The correct option is A x2+y2−2x+10y+1=0 Let r be the radius of required circle. Then r is perpendicular distance of point (1,−5) from 3x+4y−8=0 ⇒r=|3(1)+4(−5)−8|√32+42⇒r=5 Hence required equation of circle is, (x−1)2+(y+5)2=52⇒x2+y2−2x+10y+1=0

Equation of the circle, centred at $(1, - 5)$ and touching the line $3 x + 4 y = 8,$ is