Question

plese answer my question guys​

Answer 1

Let the equation of the circle be

x2+y2+2gx+2f y+λ=0.

Since it passes through (0,a) and (0,−a).

∴a2=2fa+λ=0

a2−2fa+λ=0.

Subtracting, we get 4fa=0,∴f=0; and λ=−a2.

Hence the circle becomes

x2+y2+2gx−a2=0.

Centre is (−g,0)

and radius =g2−λ

​=g2+a2

​

It is given that the circle touches the line

y=mx+c or mx−y+c=0.

Hence perpendicular from centre should be equal to radius

(m2+1)

​−mg+c​=g2+a2

​.

Square (c−mg)2=(g2+a2)(m2+1)

or g2+2gmc+{a2(1+m2)−c2}=0.

∴g1​g2​=a2(1+m2)−c2.....(1)

Above is a quadratic in g and gives us two values of g showing that there will be two circles which satisfy the given conditions. If the two values of g be g1​ and g2​ then the two circles are

x2+y2+2g1​x−a2=0

and x2+y2+2g2​x−a2=0.

These two will intersect orthogonally if

2g1​g2​=2f1​f2​=c1​+c2​

2{a2(1+m2)−c2}+0=−a2−a2=−2a2

or a2(1+m2)−c2=−a2

or a2(2+m2)=c2

is the required condition.

Answer 2

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