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Mathematics Question on Circle

Suppose the line joining distinct points P and Q on (x−2)2+(y−1)2=r2(x-2)^{2}+(y-1)^{2}=r^{2} is the diameter of (x−1)2+(y−3)2=4 (x-1)^2+(y-3)^2=4.The value of r is

A

22

B

33

C

11

D

99

E

44

Answer

22

Explanation

Solution

Given that
The line joining distinct points P and Q on the circle with the equation (x−2)2+(y−1)2=r2(x - 2)^2 + (y - 1)^2 = r^2
the diameter of the circle represented by : (x−1)2+(y−3)2=4.(x - 1)^2 + (y - 3)^2 = 4.
Since P and Q lie on the diameter of the circle (x−1)2+(y−3)2=4(x - 1)^2 + (y - 3)^2 = 4, the midpoint of the diameter will be the center of the circle.
So the diameter of the circle is (1,3).(1, 3).
Then to get the mid point of PQ we can write :
Midpoint of PQ = Center of the circle ((xp+xq)2,(yp+yq)2)=(1,3)(\dfrac{(x_p + x_q)} {2}, \dfrac{(y_p + y_q)}{2} )= (1, 3)
Distance(P, Q) =2r2r
Using the distance formula for points P and Q on the circle (x−2)2+(y−1)2=r2(x - 2)^2 + (y - 1)^2 = r^2:
Distance (P,Q)2=[(xq−xp)2+(yq−yp)2](P, Q)^2 = [(x_q - x_p)^2 + (y_q - y_p)^2]
Now, we can set up the equations as :
(xq−xp)2+(yq−yp)2=(xp−2)2+(yp−1)2=r2...(1)(x_q - x_p)^2 + (y_q - y_p)^2 = (x_p - 2)^2 + (y_p - 1)^2 = r^2 ... (1)
(xq−xp)2+(yq−yp)2=4...(2)(x_q - x_p)^2 + (y_q - y_p)^2 = 4 ... (2)
Equating both the equations we get,
r2=4r^2 = 4
⇒r=2r=2
Hence the value of rr is 22
So, the correct option is (A) : 2.