Q. 124.1( 7 Votes )

# Find the equations of the circles touching y - axis at (0, 3) and making an intercept of 8 units on the x - axis.

Answer :

Given that the circle is touching the y - axis at (0,3) and making an intercept of 8 units on the x - axis.

Let us assume the circle intersects x - axis at the points A, B. Then the length of AB = 8units.

Let us assume ‘O’ be the centre of the circle and ‘M’ be the mid - point of the line AB.

Since circle touches y - axis at (0,3), we assume the centre of the circle be (h,3).

From the figure we get OM = 3 units and AM = 4units. The points AMO forms a right angled triangle. We have AO as radius.

⇒ AO^{2} = OM^{2} + AM^{2}

⇒ AO^{2} = 3^{2} + 4^{2}

⇒ AO^{2} = 9 + 16

⇒ AO^{2} = 25

⇒

⇒ AO = 5

We have circle with centre (h,3), passing through the point (0,3) and having radius 5 units.

We know that radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

We know that the distance between the two points (x_{1},y_{1}) and (x_{2},y_{2}) is .

Let us assume r is the radius of the circle.

⇒

⇒

⇒ 5 = |h|

⇒ h = ±5 ..... (1)

We have circle with centre (±5,3) and having radius 5 units.

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±5)^{2} + (y - 3)^{2} = 5^{2}

⇒ x^{2}±10x + 25 + y^{2} - 6y + 9 = 25

⇒ x^{2} + y^{2}±10x - 6y + 9 = 0

∴The equations of the circles is x^{2} + y^{2}±10x - 6y + 9 = 0.

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