Find the equation of the circle concentric with $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ and which touches the y-axis.

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

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x^{2} + y^{2} - 4 x - 6 y + 18 = 0

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

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x^{2} + y^{2} - 4 x - 6 y + 3 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$
Center of given circle = $(\frac{4}{2}, \frac{6}{2}) = (2, 3)$

Let $(x - 2)^{2} + (y - 3)^{2} = r^{2}$ be a concentric circle

Since x = 0 is a tangent

Radius = Perpendicular distance from center to tangent

$⟹ r = \frac{| 2 + 0 |}{\sqrt{1 + 0}} ⟹ r = 2$

Equation of circle is $(x - 2)^{2} + (y - 3)^{2} = 2^{2}$

$⟹ x^{2} + y^{2} - 4 x - 6 y + 9 = 0$

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Find the equation of the circle concentric with x2+y2−4x−6y−3=0 and which touches the y-axis.

The correct option is B x2+y2−4x−6y+9=0Center of given circle = (42,62)=(2,3) Let (x−2)2+(y−3)2=r2 be a concentric circle Since x = 0 is a tangent Radius = Perpendicular distance from center to tangent ⟹r=|2+0|√1+0⟹r=2 Equation of circle is (x−2)2+(y−3)2=22 ⟹x2+y2−4x–6y+9=0

Find the equation of the circle concentric with $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ and which touches the y-axis.