Q. 94.0( 9 Votes )

# Find the equation of the circle which touches the axes and whose centre lies on x – 2y = 3.

Answer :

We need to find the equation of the circle the axes and centre lies on x - 2y = 3.

Let us assume the circle touches the axes at (a,0) and (0,a) and we get the radius to be |a|.

We get the centre of the circle as (a, a). This point lies on the line x – 2y = 3

⇒ a - 2(a) = 3

⇒ - a = 3

⇒ a = - 3

Centre = (a, a) = ( - 3, - 3) and radius of the circle(r) = | - 3| = 3

We have circle with centre ( - 3, - 3) and having radius 3.

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)^{2} + (y - q)^{2} = r^{2}

Now we substitute the corresponding values in the equation:

⇒ (x - ( - 3))^{2} + (y - ( - 3))^{2} = 3^{2}

⇒ (x + 3)^{2} + (y + 3)^{2} = 9

⇒ x^{2} + 6x + 9 + y^{2} + 6y + 9 = 9

⇒ x^{2} + y^{2} + 6x + 6y + 9 = 0.

∴The equation of the circle is x^{2} + y^{2} + 6x + 6y + 9 = 0.

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