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If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y +2p – 5= 0 and x2 + y2 +2x + 2y – p2 = 0, then there is a circle passing through P, Q, and (1, 1) for -

If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y +2p – 5= 0 and x2 + y2 +2x + 2y – p2 = 0, then there is a circle passing through P, Q, and (1, 1) for -

Grade:12

3 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
let given circles be s= x2 + y2 + 3x + 7y +2p – 5= 0 and s1=x2 + y2 +2x + 2y – p2 = 0
Equation of required circle is s+\lambdas1=0
As it passes through (1,1) the value of\lambda=\frac{-7-2p}{6-p^{2}}
if 7+2p=0 then it becomes the second circle,therefore it is true for all values of p.
Thanks and Regards
Shaik Aasif
Deepak Kumar Shringi
askIITians Faculty 4405 Points
2 years ago
Dear student please find the image below562-1503_Capture.PNG
Samyak
15 Points
one year ago
P and Q are the points of intersection of the line 
x/a + y/b=1 
 (a>O, b>O) 
with the x- and y-axes respectively. The distance PQ is 20 and the gradient of PQ is - 3. 
Find the values of a and b.

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