Question

The straight line 2x+3y-k=0 (k>0) cuts the x-and y-axis at A and B. The area of △OAB is 12sq. units, where O denotes the origin. The equation of circle having AB as diameter is?

a) x^2+ y^2-6x-4y=0
b) x^2+ y^2-6x-4y=13
c) x^2+ y^2+6x-4y=0
d) x^2+ y^2-6x+4y=0
e) x^2+ y^2+6x+4y=13

Answer 1

`3y = k - 2x`

`y = 1/3k - 2/3x`

The y-intercept is given by `y = 1/3k`. The x intercept is given by `x = 1/2k`.

The area of a triangle is given by `A = (b xx h)/2`.

`12= (1/3k xx 1/2k)/2`

`24 = 1/6k^2`

`24/(1/6) = k^2`

`144 = k^2`

`k = +-12`

We now need to determine the measure of the hypotenuse of the theoretical triangle.

`6^2 + 4^2 = c^2`

`36 + 16 = c^2`

`52 = c^2`

`sqrt(52) = c`

`2sqrt(13) = c`

The equation of the circle is given by `(x- p)^2 + (y - q)^2 = r^2`, where `(p, q)` is the centre and `r` is the radius.

The centre will occur at the mid-point of AB.

By the midpoint formula:

`m.p = ((x_1 + x_2)/2, (y_1 + y_2)/2)`

`m.p = ((6 + 0)/2, (4 + 0)/2)`

`m.p = (3, 2)`

So, the equation of the circle is `(x - 3)^2 + (y - 2)^2 = 52`

If we multiply this to the form of the choices above, we get:

`x^2 - 3x + 9 + y^2 - 4y + 4 = 52`

`x^2 - 3x +y^2 - 4y - 39 = 0`

This is none of the choices, so I've requested other contributors to check my answer.

Hopefully this helps!