Answer:
The point Q(6,7) lie outside the circle.
Step-by-step explanation:
We know that, the general equation of a circle,
---------(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](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D%20%2B2%28-2%29x%2B2%28-4%29y%2Bc%3D0%5C%5Cx%5E%7B2%7D%20%2By%5E%7B2%7D%20-4x-8y%2Bc%3D0)
The point L(0,8) is on the circle, the this point can satisfy the above equation.
So, ![0^{2} +8^{2} -4\times0-8\times8+c=0\\0+64-0-64+c=0\\c=0](https://tex.z-dn.net/?f=0%5E%7B2%7D%20%2B8%5E%7B2%7D%20-4%5Ctimes0-8%5Ctimes8%2Bc%3D0%5C%5C0%2B64-0-64%2Bc%3D0%5C%5Cc%3D0)
Now, the equation of circle is
![x^{2} +y^{2} -4x-8y=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D%20-4x-8y%3D0)
Putting the
and
in the above equation, we get
![6^{2} +7^{2} -4\times6-8\times7\\36+49-24-56\\85-80\\](https://tex.z-dn.net/?f=6%5E%7B2%7D%20%2B7%5E%7B2%7D%20-4%5Ctimes6-8%5Ctimes7%5C%5C36%2B49-24-56%5C%5C85-80%5C%5C)
5, which is greater than 0.
So, the point Q(6,7) lie outside the given circle.