Answer:
The point Q(6,7) lie outside the circle.
Step-by-step explanation:
We know that, the general equation of a circle,
[tex]x^{2} +y^{2} +2gx+2fy+c=0[/tex] ---------(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
[tex]x^{2} +y^{2} +2(-2)x+2(-4)y+c=0\\x^{2} +y^{2} -4x-8y+c=0[/tex]
The point L(0,8) is on the circle, the this point can satisfy the above equation.
So, [tex]0^{2} +8^{2} -4\times0-8\times8+c=0\\0+64-0-64+c=0\\c=0[/tex]
Now, the equation of circle is
[tex]x^{2} +y^{2} -4x-8y=0[/tex]
Putting the [tex]x=6[/tex] and [tex]y=7[/tex] in the above equation, we get
[tex]6^{2} +7^{2} -4\times6-8\times7\\36+49-24-56\\85-80\\[/tex]
5, which is greater than 0.
So, the point Q(6,7) lie outside the given circle.
