Equation of the circle which touches the lines $4 x + 3 y - 5 = 0, 4 x + 3 y + 15 = 0$ and having centre on the line $3 x + 2 y + 4 = 0$ is :

x^{2} + y^{2} + 4 x - 2 y + 1 = 0

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x^{2} + y^{2} - 4 x + 2 y + 1 = 0

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x^{2} + y^{2} + 4 x - 2 y - 11 = 0

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x^{2} + y^{2} - 4 x + 2 y - 11 = 0

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Solution

The correct option is A $x^{2} + y^{2} + 4 x - 2 y + 1 = 0$
Let $P (h, k)$ is centre of circle

$∴ r = ∣ ∣ ∣ \frac{4 h + 3 k - 5}{5} ∣ ∣ ∣$ ---(1)

and $r = ∣ ∣ ∣ \frac{4 h + 3 k + 15}{5} ∣ ∣ ∣$ --- (2)
From (1) & (2), $4 h + 3 k - 5 = - 4 h - 3 k - 15$
$\Rightarrow 4 h + 3 k + 5 = 0$ ---(3)
And $P (h, k)$ lie on $3 x + 2 y + 4 = 0$
$3 h + 2 k + 4 = 0$ ---(4)

From (3) & (4), we get, $h = - 2, k = 1$

$∴ r = ∣ ∣ ∣ \frac{- 8 + 3 - 5}{5} ∣ ∣ ∣ = 2$

So, equation of circle is

$(x + 2)^{2} + (y - 1)^{2} = (2)^{2}$
$\Rightarrow x^{2} + y^{2} + 4 x - 2 y + 1 = 0$

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Similar questions

Q. If one of the diameters of the circle

x^{2} + y^{2} - 2 x - 6 y + 6 = 0

is a chord to the circle with centre

(2, 1)

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Q. The centre of similitude of the two circles

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

and

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

Q. The radical centre of the circles

x^{2} + y^{2} - 4 x - 6 y + 5 = 0, x^{2} + y^{2} - 2 x - 4 y - 1 = 0, x^{2} + y^{2} - 6 x - 2 y = 0

lies on the line

Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

Q. A transverse common tangent to the circles

x^{2} + y^{2} + 4 x + 2 y = 4

and

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

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Equation of the circle which touches the lines 4x+3y−5=0, 4x+3y+15=0 and having centre on the line 3x+2y+4=0 is :

Equation of the circle which touches the lines $4 x + 3 y - 5 = 0, 4 x + 3 y + 15 = 0$ and having centre on the line $3 x + 2 y + 4 = 0$ is :