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in Mathematics by (67.9k points)

P is a point (a,b) in the first quadrant. If the two circles which pass throngh P and touch both the co-ordinate axes cut at right angles, then

(a)  a– 6ab + b2 = 0

(b)  a+ 2ab – b2 = 0

(c)  a– 4ab + b2 = 0

(d)  a– 8ab + b2  = 0 

1 Answer

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Best answer

Correct option   (c) a– 4ab + b2 = 0

Explanation:

Equation of the two circles be

(x – r)+ (y – r)2 = r⇒ x+ y– 2xr – 2yr + r2 =0 

These two circles passes through (a,b)

(a – r)+ (b – r)2 = r2

a+ r– 2ar + b+ r– 2br – r2 = 0

r2– 2r(a + b) +(a+ b2 ) = 0

It is a quadratic equation in r 

r1 + r2 = 2(a + b)and r1 r2 = a+ b2

Condition for orthogonality is

2g1 g2 +2f1f2 = C1 + C2

2r1r2 + 2r1 r = r1 + r22 

4r1 r2 = r1+ r22 

6r1r2 =  r1+ r2+ 2r1r2

6r1r2 = (r1  + r2)2

6(a+ b2) = 4 (a + b)2

6a2  + 6b= 4a+ 4b+ 8ab

2a+ 2b– 8ab = 0

a+ b– 4ab = 0

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