Find the equation of the circle passing the point 1-2 and touching the line 2 x-y-1-0 at the point 1-1-A- x2-y2-8 x-y-y-0B- x2-y2-4 x-2 y-5-0- x 2- y 2-8 x - y -5-0D- x 2- y 2 4 x -2 y -5-0

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Equation of Family of Circles Touching a Line and Passing through a Given Point on the Line

Find the equa...

Question

Find the equation of the circle passing the point (1, 2) and touching the line 2x + y - 1 = 0 at the point (1, -1).

$x^{2} + y^{2} + 8 x + y + y = 0$

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

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$x^{2} + y^{2} - 8 x - y + 5 = 0$

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

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Solution

The correct option is C
$x^{2} + y^{2} - 8 x - y + 5 = 0$

Here, equation of family of circle is

$s + λ L = 0$

$(x - 1)^{2} + (y + 1)^{2} + λ (2 x + y - 1) = 0$ .........(1)

Since, It passes through the point (1,2)

Then

$(1 - 1)^{2} + (2 + 1)^{2} + λ (2 \times 1 + 2 - 1) = 0$

$9 + λ (3) = 0$

$λ = \frac{- 9}{3} = - 3$

Substituting the value of $λ$ in equation (1)

$(x - 1)^{2} + (y + 1)^{2} + (- 3) (2 x + y - 1) = 0$

$x^{2} + y^{2} - 8 x - y + 5 = 0$

Equation of circle is

$x^{2} + y^{2} - 8 x - y + 5 = 0$

Option c is correct

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