Q. 7

# Find the equation of the circle whose diameter is the line segment joining (- 4, 3) and (12, - 1). Find also the intercept made by it on the y-axis.

Answer :

Given that we need to find the equation of the circle whose end points of a diameter are (- 4,3) and (12, - 1).

We know that the centre is the mid - point of the diameter.

⇒ Centre(C) =

⇒ C = (4,1)

We have a circle with centre (4,1) and passing through the point (- 4,3).

We know that the 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.

⇒

⇒

⇒

⇒

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^{2} - 8x + 16 + y^{2} - 2y + 1 = 68

⇒ x^{2} + y^{2} - 8x - 2y - 51 = 0 ..... (1)

To find the y - intercept, we need the find the points at which the circle intersects the y - axis.

We know that x = 0 on y - axis. Substituting x = 0 in (1) we get

⇒ 0^{2} + y^{2} - 8(0) - 2y - 51 = 0

⇒ y^{2} - 2y - 51 = 0

⇒

⇒

⇒ y = 1 ± 2√13

∴The y - intercepts are 1 ± 2√13.

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