Question

A circle is centered at R(2,4)R(2,4)R, (, 2, comma, 4, ). The point L(0,8)L(0,8)L, (, 0, comma, 8, )is on the circle. Where does the point Q(6,7)Q(6,7)Q, (, 6, comma, 7, )lie?

Answer 1

The point Q(6,7) lie outside the circle.

Explanation:

We know that, the general equation of a circle,

`x^(2) +y^(2) +2gx+2fy+c=0` ---------(1)

Centre is (-g, -f) = (2,4) (given)

So, -g = 2⇒g = -2

and, -f = 4⇒f = -4

Putting the value of g and f in equation(1), we get

`x^(2) +y^(2) +2(-2)x+2(-4)y+c=0\\x^(2) +y^(2) -4x-8y+c=0`

The point L(0,8) is on the circle, the this point can satisfy the above equation.

So,
`0^(2) +8^(2) -4*0-8*8+c=0\\0+64-0-64+c=0\\c=0`

Now, the equation of circle is

`x^(2) +y^(2) -4x-8y=0`

Putting the
`x=6` and
`y=7` in the above equation, we get

`6^(2) +7^(2) -4*6-8*7\\36+49-24-56\\85-80\\`

5, which is greater than 0.

So, the point Q(6,7) lie outside the given circle.