# Discuss the position of the points $\left[ {1,2} \right]$ and $\left[ {6,0} \right]$ with respect to the circle ${x^2} + {y^2} - 4x + 2y - 11 = 0$.

Last updated date: 20th Mar 2023

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Answer

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Hint- Here, we will proceed by using the formula for the position of a point with respect to a given circle.

Given, equation of circle is ${x^2} + {y^2} - 4x + 2y - 11 = 0$

Let the given points be ${\text{P}}\left[ {1,2} \right]$ and ${\text{Q}}\left[ {6,0} \right]$.

As we know that any point ${\text{A}}\left[ {a,b} \right]$ lies inside, on or outside the circle ${\text{S: }}{x^2} + {y^2} + 2gx + 2fy + c = 0$ according to as ${{\text{S}}_1}$ is less than or equal to or greater than zero respectively where ${{\text{S}}_1} = {\left( a \right)^2} + {\left( b \right)^2} + 2g\left( a \right) + 2f\left( b \right) + c$.

For point ${\text{P}}\left[ {1,2} \right]$, ${{\text{S}}_1} = {1^2} + {2^2} - 4 \times 1 + 2 \times 2 - 11 = 1 + 4 - 4 + 4 - 11 = - 6 < 0$

Since, ${{\text{S}}_1}$ is less than zero which means that point ${\text{P}}\left[ {1,2} \right]$ lies inside the circle.

For point ${\text{Q}}\left[ {6,0} \right]$, ${{\text{S}}_2} = {6^2} + 0 - 4 \times 6 + 2 \times 0 - 11 = 36 - 24 - 11 = 1 > 0$

Since, ${{\text{S}}_2}$ is greater than zero which means that point ${\text{Q}}\left[ {6,0} \right]$ lies outside the circle.

Note- In these types of problems, simply substitute the x and y coordinates of the point in the LHS of the given equation of circle provided the RHS of the given equation of circle is zero and determine the value obtained. Then, finally by comparing it with the formula we get to know whether the given point lies inside, on or outside of the circle.

Given, equation of circle is ${x^2} + {y^2} - 4x + 2y - 11 = 0$

Let the given points be ${\text{P}}\left[ {1,2} \right]$ and ${\text{Q}}\left[ {6,0} \right]$.

As we know that any point ${\text{A}}\left[ {a,b} \right]$ lies inside, on or outside the circle ${\text{S: }}{x^2} + {y^2} + 2gx + 2fy + c = 0$ according to as ${{\text{S}}_1}$ is less than or equal to or greater than zero respectively where ${{\text{S}}_1} = {\left( a \right)^2} + {\left( b \right)^2} + 2g\left( a \right) + 2f\left( b \right) + c$.

For point ${\text{P}}\left[ {1,2} \right]$, ${{\text{S}}_1} = {1^2} + {2^2} - 4 \times 1 + 2 \times 2 - 11 = 1 + 4 - 4 + 4 - 11 = - 6 < 0$

Since, ${{\text{S}}_1}$ is less than zero which means that point ${\text{P}}\left[ {1,2} \right]$ lies inside the circle.

For point ${\text{Q}}\left[ {6,0} \right]$, ${{\text{S}}_2} = {6^2} + 0 - 4 \times 6 + 2 \times 0 - 11 = 36 - 24 - 11 = 1 > 0$

Since, ${{\text{S}}_2}$ is greater than zero which means that point ${\text{Q}}\left[ {6,0} \right]$ lies outside the circle.

Note- In these types of problems, simply substitute the x and y coordinates of the point in the LHS of the given equation of circle provided the RHS of the given equation of circle is zero and determine the value obtained. Then, finally by comparing it with the formula we get to know whether the given point lies inside, on or outside of the circle.

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