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# The center of the circle is $\left( {x + 2,x - 1} \right)$. Find $x$ if the circle passes through $\left( {2, - 2} \right)\& \left( {8, - 2} \right)$.

Last updated date: 20th Jun 2024
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Hint: Here, we have to find the value of the variable $x$. We have $\left( {x + 2,x - 1} \right)$ as the center of circle and two points on the circumference of the circle are given as $\left( {2, - 2} \right)\& \left( {8, - 2} \right)$.we know the distance of the center of the circle from any point on the circumference of the circle is equal to radius. Thus, the distance of the point $\left( {x + 2,x - 1} \right)$ from the two given points $\left( {2, - 2} \right)\& \left( {8, - 2} \right)$ is equal.

Complete Complete Step by Step Solution:
First we will draw the diagram of the circle as per the given information.

Given that, a circle passes through points $\left( {2, - 2} \right)\& \left( {8, - 2} \right)$.
This means that these points lie on the circumference of the circle.
We know that the distance from the center of the circle to any point on the circumference is equal to each other as it is called as radius.
So the distance of both the points from the center is equal.
Now, we will calculate the distance between the point $\left( {x + 2,x - 1} \right)$ and $\left( {2, - 2} \right)$ using the distance formula.
Distance of the point $\left( {x + 2,x - 1} \right)$and $\left( {2, - 2} \right)$ $= \sqrt {\left[ {{{\left( {x + 2 - 2} \right)}^2} + {{\left( {x - 1 + 2} \right)}^2}} \right]}$
On further simplification, we get
$\Rightarrow$ Distance of the point $\left( {x + 2,x - 1} \right)$ and $\left( {2, - 2} \right)$ $= \sqrt {\left[ {{x^2} + {{\left( {x + 1} \right)}^2}} \right]}$……………….$\left( 1 \right)$
Now, we will calculate the distance between the point $\left( {x + 2,x - 1} \right)$ and $\left( {8, - 2} \right)$ using the distance formula.
Distance of the point $\left( {x + 2,x - 1} \right)$and $\left( {8, - 2} \right)$ $= \sqrt {\left[ {{{\left( {x + 2 - 8} \right)}^2} + {{\left( {x - 1 + 2} \right)}^2}} \right]}$
On further simplification, we get
$\Rightarrow$ Distance of the point $\left( {x + 2,x - 1} \right)$ and $\left( {8, - 2} \right)$ $= \sqrt {\left[ {{{\left( {x - 6} \right)}^2} + {{\left( {x + 1} \right)}^2}} \right]}$……………….$\left( 2 \right)$
Equating equation 1 and equation 2 as both are the radius of the same circle, we get
$\sqrt {\left[ {{x^2} + {{\left( {x + 1} \right)}^2}} \right]} = \sqrt {\left[ {{{\left( {x - 6} \right)}^2} + {{\left( {x + 1} \right)}^2}} \right]}$
Squaring on both sides, we cancel the roots. Hence we get,
$\Rightarrow {x^2} + {\left( {x + 1} \right)^2} = {\left( {x - 6} \right)^2} + {\left( {x + 1} \right)^2}$
On simplifying the terms, we get
$\Rightarrow {x^2} = {\left( {x - 6} \right)^2}$
Applying exponents on the bases, we get
$\Rightarrow {x^2} = {x^2} - 36 + 12x$
Rearranging the terms, we get
$\begin{array}{l} \Rightarrow 12x = 36\\ \Rightarrow x = 3\end{array}$

Hence, the value of x is 3.

Note:
1) The formula which we have used here for finding the distance between the two points is in two dimensional.
2) The distance formula for three dimensional is different.
3) Say we have to find the distance between the points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$
4) So the distance between these two points is
$\sqrt {\left[ {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{({z_2} - {z_1})}^2}} \right]}$