Question

Discuss the position of the points (1,2) and (6,0) with respect to the circle $x^{2}+y^{2}-4x+2y-11=10$.

Answer 1

Hint: In this question it is given that we have to find the position of the points (1,2) and (6,0) with respect to the circle $x^{2}+y^{2}-4x+2y-11=10$. So to find the solution we have to determine these given two points lies in which portion of the circle.
Let (a,b) be any point,
If, $a^{2}+b^{2}-4a+2b-11<10$, then (a,b) lies inside the circle.
If, $a^{2}+b^{2}-4a+2b-11=10$, then (a,b) lies on the circle.
If, $a^{2}+b^{2}-4a+2b-11>10$, then (a,b) lies outside the circle.

Complete step-by-step answer:
First of all we check for the point (1,2), so by putting the point in the left hand side of the equation, we get,
$x^{2}+y^{2}-4x+2y-11$
=$1^{2}+2^{2}-4\times 1+2\times 2-11$
=1+4-4+4-11 = -6, which is less than 10.
i.e, $1^{2}+2^{2}-4\times 1+2\times 2-11$<10.
So we can say that (1,2) lies inside the circle.

Now for the point (6,0)
$x^{2}+y^{2}-4x+2y-11$
=$6^{2}+0^{2}-4\times 6+2\times 0-11$
=36-24-11 = 1 which is also less than 10.
So this point also lies inside the circle.

Note: While solving this type of question you need to keep in mind, in order to find the solution you have to put the points on the left hand side of the given equation. If the left hand side(LHS) value is greater than the right hand side(RHS) then we can say that the points lie outside the circle and if equal to the RHS value then the points lie on the circle and if less than RHS then the point must be inside the circle.