The equation of the circle having its centre | vedclass

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(D) The given lines $3x + 4y - 18 = 0$ and $3x + 4y + 2 = 0$ are parallel tangents to the circle.The distance between these parallel lines is $d = \fr

Vedclass · Accepted Answer

$x^2 + y^2 + 8x - 10y + 37 = 0$

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(D) The given lines $3x + 4y - 18 = 0$ and $3x + 4y + 2 = 0$ are parallel tangents to the circle.The distance between these parallel lines is $d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} = \frac{|2 - (-18)|}{\sqrt{3^2 + 4^2}} = \frac{20}{5} = 4$.The diameter of the circle is $4$,so the radius $r = \frac{4}{2} = 2$.Let the centre of the circle be $(h, k)$. Since the centre lies on $2x + y + 3 = 0$,we have $2h + k + 3 = 0 \Rightarrow k = -2h - 3$.The perpendicular distance from $(h, k)$ to the tangent $3x + 4y + 2 = 0$ is equal to the radius $r = 2$:$\frac{|3h + 4k + 2|}{\sqrt{3^2 + 4^2}} = 2$ $\Rightarrow |3h + 4(-2h - 3) + 2| = 10$ $\Rightarrow |3h - 8h - 12 + 2| = 10$ $\Rightarrow |-5h - 10| = 10$.This gives $-5h - 10 = 10 \Rightarrow h = -4$ or $-5h - 10 = -10 \Rightarrow h = 0$.If $h = -4$,then $k = -2(-4) - 3 = 5$. The centre is $(-4, 5)$.The equation of the circle is $(x + 4)^2 + (y - 5)^2 = 2^2$ $\Rightarrow x^2 + 8x + 16 + y^2 - 10y + 25 = 4$ $\Rightarrow x^2 + y^2 + 8x - 10y + 37 = 0$.

The equation of the circle having its centre on the line $2x + y + 3 = 0$ and having the lines $3x + 4y - 18 = 0$ and $3x + 4y + 2 = 0$ as tangents is:

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