Question

The centre of a circle is \[(x + 2,{\text{ }}x - 1).\]Find x if the circle passes through \[(2, - 2){\text{ }}\& {\text{ }}(8, - 2).\]

Answer 1

Hint: The centre of a circle is given by \[(x + 2,{\text{ }}x - 1).\]
Since A \[(2, - 2){\text{ }}\] and B \[(8, - 2)\] are two point on the circle, therefore their distances from the centre will be equal (i.e. the radius of the circle)
i.e. $OA = OB = r$
Therefore find OA and OB and equate them to solve for x.
We can also find the value of x by using $\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)$


Complete step by step answer:

The centre of a circle is given by \[(x + 2,{\text{ }}x - 1).\]
Also, the circle passes through \[(2, - 2){\text{ }}\& {\text{ }}(8, - 2).\]
The coordinate of centre of circle is given by O\[(x + 2,{\text{ }}x - 1).\]
Since A \[(2, - 2){\text{ }}\] and B \[(8, - 2)\] are two point on the circle, therefore their distances from the centre will be equal (i.e. the radius of the circle)
i.e. $OA = OB = r$
We use distance formula to calculate distance between 2 points, if given two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\], so distance between them is given by $D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Now,
Now, on applying distance formula on OA we get,
$OA = \sqrt {{{\left( {x + 2 - 2} \right)}^2} + {{\left( {x - 1 + 2} \right)}^2}} $
On simplification we get,
$ \Rightarrow OA = \sqrt {{x^2} + {{\left( {x + 1} \right)}^2}} $
On expanding the square we get,
$ \Rightarrow OA = \sqrt {{x^2} + {x^2} + 2x + 1} $
On simplification we get,
$ \Rightarrow OA = \sqrt {2{x^2} + 2x + 1} $
Again, on applying distance formula on OB we get,
 $OB = \sqrt {{{\left( {x + 2 - 8} \right)}^2} + {{\left( {x - 1 + 2} \right)}^2}} $
On simplification we get,
$ \Rightarrow OB = \sqrt {{{\left( {x - 6} \right)}^2} + {{\left( {x + 1} \right)}^2}} $
On expanding the square we get,
$ \Rightarrow OB = \sqrt {{x^2} - 12x + 36 + {x^2} + 2x + 1} $
On further simplification we get,
$ \Rightarrow OB = \sqrt {2{x^2} - 10x + 37} $
Now, OA=OB
Therefore,
$\sqrt {2{x^2} + 2x + 1} = \sqrt {2{x^2} - 10x + 37} $
On squaring both the sides we get,
$ \Rightarrow 2{x^2} + 2x + 1 = 2{x^2} - 10x + 37$
On cancelling common terms we get,
$ \Rightarrow 2x + 1 = - 10x + 37$
On taking like terms together we get,
$ \Rightarrow 2x + 10x = 37 - 1$
On further simplification we get,
$ \Rightarrow 12x = 36$
$ \Rightarrow x = 3$
Hence the value of x is 3, and the coordinates of the centre of the circle is \[\left( {3 + 2,{\text{ }}3 - 1} \right) = \left( {5,{\text{ }}2} \right).\]

Note: Given points \[(2, - 2){\text{ }}\& {\text{ }}(8, - 2)\], lie on a circle, as we can see that the y coordinate of both lines are same, so we can say that the points lie on a line. So, we can say that the points \[(2, - 2){\text{ }}\& {\text{ }}(8, - 2)\] are the end points of diameter of a circle. So, the centre will be midpoint of \[(2, - 2){\text{ }}\& {\text{ }}(8, - 2)\].
So using midpoint formula which is $\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)$
So the x coordinate is $ = \dfrac{{\left( {2 + 8} \right)}}{2} = \dfrac{{10}}{2} = 5$
Also as the x coordinate of centre of circle is \[x + 2\]
On comparing we get,
\[x + 2 = 5\]
On simplification we get,
\[ \Rightarrow x = 3\]