Question

If $P\left( {2,8} \right)$ is an interior point of a circle ${x^2} + {y^2} - 2x + 4y - p = 0$ which neither touches nor intersects the axes, then set for $p$ is
A) $p < \- 1$
B) $p < \- 2$
C) $p > 96$
D) $\phi$

Answer 1

Here we will find out the coordinates of the center of the circle by simplifying and solving the equation of the circle. We will then find the radius of the circle from the in terms of $p$. Then we will use the condition of being point P lies inside the circle and the distance between the center of the circle and point P is less than the radius of the circle. Again by solving further we will get the required set for $p$.

Formula used:
We will use the formula of the distance between two points is $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$.

Complete step by step solution:
We have to first find out the coordinates of the center of the circle by simplifying and solving the equation of the circle.
Rewriting the given equation, we get
${x^2} + 1 - 2x + {y^2} + 4 + 4y - p - 5 = 0$
The above equation is now of the form of the algebraic identity, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
Now rewriting the above equation, we get
$\Rightarrow {(x - 1)^2} + {(y + 2)^2} - p - 5 = 0$
$\Rightarrow {(x - 1)^2} + {(y + 2)^2} = p + 5$
We can write the right side of the equation as
$\Rightarrow {(x - 1)^2} + {(y + 2)^2} = {\left( {\sqrt {5 + p} } \right)^2}$
Now by comparing the above equation of the circle with the basic equation of the circle i.e. ${(x - {x_0})^2} + {(y - {y_0})^2} = {r^2}$ where, ${x_0}$ is the $x$ coordinate of the center of the circle and $r$ is the radius of the circle and ${y_0}$ is the $y$ coordinate of the center of the circle.
Therefore from the comparison, we can say that the coordinates of the center of a given circle is $\left( {1, - 2} \right)$ and radius of the circle is $\sqrt {5 + p}$.
It is given that the point $P(2,8)$ is an interior point of a circle. Therefore from this we get that the distance between the center of the circle and point P is less than the radius of the circle.
$\Rightarrow \sqrt {{{(2 - 1)}^2} + {{(8 + 2)}^2}} < \sqrt {5 + p}$
Simplifying the above equation, we get
$\Rightarrow \sqrt {1 + 100} < \sqrt {5 + p}$
$\Rightarrow \sqrt {101} < \sqrt {5 + p}$
Now removing the square root from the both side of the equation, we get
$\Rightarrow 101 < 5 + p$
$\Rightarrow p > 101 - 5$
$\Rightarrow p > 96$

Hence, option C is the correct option.

Note:
In this question, we have transformed the given equation of the circle into the general equation of the circle. For that we have used the algebraic identities, so it becomes important for us to know how to rearrange terms so that it can be transformed into any of the algebraic identities.
In addition to this, for any point which is termed as the interior point of a circle satisfies the condition of the distance between the center of the circle and point is less than the radius of the circle.