Question

Without drawing the figure determine whether the points (0,0), (-2,1), (4,-3), (2,-6), (0,-1) lie outside or on or inside the circle $x^{2}+y^{2}-5x+2y-5=0$.

Answer 1

Hint: In this question it is given that without drawing the figure we have to determine whether the points (0,0), (-2,1), (4,-3), (2,-6), (0,-1) lie outside or on or inside the circle $x^{2}+y^{2}-5x+2y-5=0$. 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, $a^{2}+b^{2}-5a+2b-5<0$, then (a,b) lies inside the circle.
If, $a^{2}+b^{2}-5a+2b-5=0$, then (a,b) lies on the circle.
If $a^{2}+b^{2}-5a+2b-5>0$, then (a,b) lies outside the circle.

Complete step-by-step answer:
i. First of all we check for the point (0,0)
$x^{2}+y^{2}-5x+2y-5$
=$0^{2}+0^{2}-5\times 0+2\times 0-5$
= -5, which is less than 0.
i.e, $0^{2}+0^{2}-5\times 0+2\times 0-5$<0.
So we can say that (0,0) lies inside the circle.

ii. Now for the point (-2,1)
$x^{2}+y^{2}-5x+2y-5$
=$\left( -2\right)^{2} +1^{2}-5\left( -2\right) +2\times 1-5$
=4+1+10+2-5 = 12 which is greater than 0.
i.e, $\left( -2\right)^{2} +1^{2}-5\left( -2\right) +2\times 1-5$>0.
So this point lies outside the circle.

iii. Now check for (4,-3)
$x^{2}+y^{2}-5x+2y-5$
=$4^{2}+\left( -3\right)^{2} -5\times 4+2\left( -3\right) -5$
=16+9-20-6-5 = -6 <0,
(4,-3) lie inside the circle.

iv. Now the fourth point (2,-6)
$x^{2}+y^{2}-5x+2y-5$
=$2^{2}+\left( -6\right)^{2} -5\times 2+2\left( -6\right) -5$
=4+36-10-12-5 = 13 >0.
Therefore, (2,-6) lie outside the circle.

v. Now checking for the last point (0,-1)
$x^{2}+y^{2}-5x+2y-5$
=$0^{2}+\left( -1\right)^{2} -5\times 0+2\left( -1\right) -5$
=1-2-5 =4 >0,
So we can say the point (0,-1) lies outside 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.