Question

A circle passing through $(0,0),(2,6),(6,2)$ cuts the x-axis at the point ${\text{P}} \ne {\text{(0,0)}}{\text{.}}$
Then the length of ${\text{OP}}$ , where ${\text{O}}$ is the origin, is
${\text{A}}{\text{. }}\dfrac{5}{2} \\\ {\text{B}}{\text{. }}\dfrac{5}{{\sqrt 2 }} \\\ {\text{C}}{\text{. 5}} \\\ {\text{D}}{\text{. 10}} \\\$

Answer 1

-In this question firstly you have to find the unknowns in the general formula of circle
${x^2} + {y^2} + 2gx + 2fy + c = 0$ by putting the different coordinates and then resolve it into equation where $g,f,c$ are known .Then you put $y = 0$ to get points where $x - {\text{axis}}$ cut the circle. And then
Compute distance between those points.

Complete step-by-step answer:
Let ${\text{O(0,0) , Q(2,6) , R(6,2)}}$ are three given points from which a circle passes
Let the general form of required circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0{\text{ }}$ -eq.1

According to the question above, the equation of circle passes through the three given points ${\text{O,Q,R}}$ .
It means they satisfy the equation of the circle. On putting values of coordinates of ${\text{O,Q,R}}$ we get
On putting $x = 0,y = 0$ in eq.1 we get
$c = 0{\text{ }}$ -eq.2
Then eq.1 is reduced to
$\Rightarrow {x^2} + {y^2} + 2gx + 2fy = 0{\text{ }}$ --eq.3
On putting $x = 2,y = 6$ in eq.3 we get
$\Rightarrow {\text{ }}{{\text{2}}^2} + {6^2} + (2g \times 2) + (2f \times 6) = 0 \\\ \Rightarrow 4 + 36 + 4g + 12f = 0 \\\ \Rightarrow 40 + 4g + 12f = 0 \\\ \\\$
On taking “4” common from above equation we get
$\Rightarrow {\text{10 + }}g + 3f = 0{\text{ }}$ --eq.4

Now, on putting $x = 6,y = 2$ in eq.3 we get
$\Rightarrow {\text{ }}{6^2} + {2^2} + (2g \times 6) + (2f \times 2) = 0 \\\ \Rightarrow 36 + 4 + 12g + 4f = 0 \\\ \Rightarrow 40 + 12g + 4f = 0 \\\ \\\$
On taking “4” common from above equation we get
$\Rightarrow {\text{10 + 3}}g + f = 0{\text{ }}$ --eq.5
On subtracting eq.4 and eq.5 we get
$g = f{\text{ }}$ - eq.6
Put $g = f{\text{ }}$ in eq. 5 we get
$f = \dfrac{{ - 5}}{2}$
from eq. 6 we get
$g = f = \dfrac{{ - 5}}{2}{\text{ }}$ --eq.7
Put values of $g,f$ from eq.7 in eq.3 we get
$2{x^2} + 2{y^2} - 10x - 10y = 0{\text{ }}$ -eq.8
Eq.8 is our equation of a given circle.
Now, to get points where x-axis cut the circle, put $y = 0$ in eq.8
We get
$\Rightarrow 2{x^2} - 10x = 0 \\\ \Rightarrow 2x(x - 5) = 0 \\\ \Rightarrow x = 0,5 \\\$
Hence points where x-axis cut the circle are $(0,0){\text{ }}$ and $(5,0)$
Since it is given that we have to assume O to be origin $(0,0)$ then P is $(5,0)$
Then length of OP is given by the formula $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
Where $\left( {{x_1},{y_1}} \right)$ are coordinates of one point and $\left( {{x_2},{y_2}} \right)$ are the coordinates of other point.
Length of OP = $\sqrt {{{(5 - 0)}^2} + 0}$
Length of OP =5
Hence the option ${\text{C}}$ is correct.

Note: Whenever you get this type of question you have to knowledge about general equation of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and interpret of variables used in this and the formula of distance between two points which is $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ .Then resolve general equation of circle in such a way that there will no remain any unknown values.