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

Hint: -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.