Points P (4,1) , Q(6,5) , R(2,7) lie on a circle . What is the perpendicular distance of the chords PQ , QR and PR from the centre ??

Dear Student,

Please find below the solution to the asked query:

We form our diagram from given information , As :

Here , OA $\perp$ PQ  , OB $\perp$ QR and OC $\perp$ PR and we know perpendicular from center to any chord also bisects the chord , So

A , B and C are mid points of PQ , QR and PR respectively .

We know formula for mid point when coordinate of two end points given : $\left(x , y \right) = \left(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2}\right)$

So,

Coordinate of A = $\left(\frac{4 + 6}{2} , \frac{1 + 5}{2}\right) = \left(\frac{10}{2} , \frac{6}{2}\right) = \left(\mathbf{ }\mathbf{5}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{3}\mathbf{ }\right)$ ,

Coordinate of B = $\left(\frac{6 + 2}{2} , \frac{5 + 7}{2}\right) = \left(\frac{8}{2} , \frac{12}{2}\right) = \left(\mathbf{ }\mathbf{4}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{6}\mathbf{ }\right)$

And

Coordinate of C = $\left(\frac{4 + 2}{2} , \frac{1 + 7}{2}\right) = \left(\frac{6}{2} , \frac{8}{2}\right) = \left(\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{4}\mathbf{ }\right)$

We know slope of equation when coordinate of two end points given : $m = \left(\frac{y_2 - y_1}{x_2 - x_1}\right)$
So,

Slope of line PQ = $m_1 = \left(\frac{5 - 1}{6 - 4}\right)= \left(\frac{4}{2}\right) = \mathbf{2}$

And we know product of slope of two perpendicular lines will give  = - 1  , If we assume slope of line OA = m₂ , So

m₁ m₂ =  -  1  , Substitute value of ' m₁ '  as we calculated above and get :

( 2 ) m₂ = - 1 ,

m₂  = $-\frac{1}{2}$

We know equation of line when we know slope of line and a point of coordinate where that passing through :

( y -  y₁ ) = m ( x  - x₁ )

Then for line OA :  x₁ = 5 ,  y₁ = 3 and m  = $-\frac{1}{2}$ , So

( y - 3 ) = $-\frac{1}{2}$ ( x  - 5 ) ,

2 y  - 6 = - x  + 5  ,

x + 2 y  =  11                                                                              --- ( 1 )

And Slope of line QR = $m_1 = \left(\frac{7 - 5}{2 - 6}\right)= \left(\frac{2}{- 4}\right) = -\frac{\mathbf{1}}{\mathbf{2}}$

And we know product of slope of two perpendicular lines will give  = - 1  , If we assume slope of line OB = m₂ , So

m₁ m₂ =  -  1  , Substitute value of ' m₁ '  as we calculated above and get :

( $-\frac{1}{2}$ ) m₂ = - 1 ,

m₂  = 2

Then for line OB :  x₁ = 4 ,  y₁ = 6 and m  = 2 , So

( y - 6 ) = 2 ( x  - 4 ) ,

y  - 6 = 2 x  - 8  ,

2x - y  =  2                                                                                --- ( 2 )

Now we multiply by 2 in equation 2 and get :

4x - 2 y  =  4                                                                             --- ( 3 )

Now we add equation 1 and 3 and get :

5 x  =  15  ,

x  =  3 , Substitute that value in equation 1 and get

3  + 2 y  =  11  ,

2 y  = 8   ,

y  = 4 

So,

Coordinate of center  =  (  3 , 4 )

We know distance formula :  d  = $\sqrt{{\left(x_2 - x_1\right)}^2 + {\left(y_2 - y_1\right)}^2}$

For perpendicular distance of the chords PQ from center , x₁ = 3 ,  y₁ =  4 and x₂ = 5 ,  y₂ = 3 , So

OA  = $\sqrt{{\left(5 - 3\right)}^2 + {\left(3 - 4\right)}^2} = \sqrt{2^2 + {\left(- 1\right)}^2} = \sqrt{4 + 1} = \sqrt{\mathbf{5}}\mathbf{ }\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

And

For perpendicular distance of the chords QR from center , x₁ = 3 ,  y₁ =  4 and x₂ = 4 ,  y₂ = 6 , So

OB  = $\sqrt{{\left(6 - 3\right)}^2 + {\left(4 - 4\right)}^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = \mathbf{3}\mathbf{ }\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

And

For perpendicular distance of the chords PR from center , x₁ = 3 ,  y₁ =  4 and x₂ = 3 ,  y₂ = 4 , So

OC  = $\sqrt{{\left(3 - 3\right)}^2 + {\left(4 - 4\right)}^2} = \sqrt{0^2 + 0^2} = \mathbf{0}\mathbf{ }\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$


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