Find the equation of the circle which touches the axes and whose centr

1.	Find the equation of the circle which touches the axes and whose centre lies on x – 2y = 3.
Answer» 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 = r² Now by substituting the values in the equation, we get (x – (-3))² + (y – (-3))² = 3² (x + 3)² + (y + 3)² = 9 x² + 6x + 9 + y² + 6y + 9 = 9 x² + y² + 6x + 6y + 9 = 0 ∴ The equation of the circle is x² + y² + 6x + 6y + 9 = 0.

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

Answer»

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 = r²

Now by substituting the values in the equation, we get

(x – (-3))² + (y – (-3))² = 3²

(x + 3)² + (y + 3)² = 9

x² + 6x + 9 + y² + 6y + 9 = 9

x² + y² + 6x + 6y + 9 = 0

∴ The equation of the circle is x² + y² + 6x + 6y + 9 = 0.

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