Question

Find the position of the points (1, 2) and (6, 0) w.r.t the circle ${{\text{x}}^2} + {{\text{y}}^2} - 4x + 2y - 11 = 0$.

Answer 1

Hint: Let us put the coordinates of two given points in the equation of the circle. So, that we can get whether these points are inside or outside the given circle.

Complete step-by-step answer:
As we know that position of point with respect to a circle means whether the point is inside, outside or on the given circle.
As we know that according to the properties of circle if ${\text{a}}{{\text{x}}^2} + {\text{b}}{{\text{y}}^2} + 2{\text{gx}} + 2{\text{fy}} + {\text{c}} = 0$ is the equation of any circle where a, b, c, g and f can be any constant. Then S(h, k) gives us the position of point (h, k) with respect to the circle ${\text{a}}{{\text{x}}^2} + {\text{b}}{{\text{y}}^2} + 2{\text{gx}} + 2{\text{fy}} + {\text{c}} = 0$.
If S(h, k) > 0 then the point (h, k) is outside the given circle.
If S(h, k) < 0 then the point (h, k) is inside the given circle.
And if S(h, k) = 0 then the point (h, k) is on the given circle.
And to calculate the value of S(h, k) we put x = h and y = k in the given equation of circle.
So, let us find the position of point (1, 2) with respect to the circle ${{\text{x}}^2} + {{\text{y}}^2} - 4x + 2y - 11 = 0$
S(1, 2) = ${\left( 1 \right)^2} + {\left( 2 \right)^2} - 4*1 + 2*2 - 11$ = – 6
So, S(1, 2) < 0
Hence the point (1, 2) is inside the circle ${{\text{x}}^2} + {{\text{y}}^2} - 4x + 2y - 11 = 0$.
Now, let us find the position of point (6, 0) with respect to the circle ${{\text{x}}^2} + {{\text{y}}^2} - 4x + 2y - 11 = 0$
S(1, 2) = ${\left( 6 \right)^2} + {\left( 0 \right)^2} - 4*6 + 2*0 - 11$ = 1
So, S(6, 0) > 0
Hence the point (6, 0) is outside the circle ${{\text{x}}^2} + {{\text{y}}^2} - 4x + 2y - 11 = 0$.

Note: Whenever we come up with this type of problem then we have to find S(h, k) for the given circle where (h, k) is the point whose position with respect to the circle we had to find. And to find the value of S(h, k) we put x = h and y = k in the given equation of the circle. If S(h, k) > 0 then the point (h, k) is outside, if S(h, k) < 0 then the point (h, k) is inside and if S(h, k) = 0 then the point (h, k) is on the given circle. So, this will be the efficient way to find the position of the point with respect to the circle.