Question

koi hai jo is question ko solve krde​

Answer 1

Hey mate❤

given equation of circle

x²+y²+2ax+c=0. ...(1)

x²+y²+2bx+c=0. ...(2)

compare by general equation of circle

X²+Y²+2GX+2FY+C=0

SO,

G1=a, F1=0 ,C1=c

and,

G2=b,F2=0,C2=c

Now,

centre of first circle

C'= (-G1,-F1)

or,

C'=(-a,0)

centre of second circle

c"=(-G2,-F2)

C"=(-b,0)

so,

distance between to circle

will be

C'C"=√(a²+b²)

Radius of first circle

r1=(√(g²+f²-c)

r1=√(a²-c)

radius of second circle

r2= √(b²-c)

so,

if two circle touch each other

c'c"= r1+r2

=>√(a²+b²)=√(a2-c)+√(b²-c)

squaring both side

=>(a2+b²)=(a²-c)+(b²-c)+2√(a²-c)× √(b²-c)

=>2√(a²-c)× √(b²-c)=2c

again squaring both side

we get ,

=>4(a²-c)(b²-c)=4c²

=>a²b²-b²c-a²c+c²=c²

=>a²b²=b²c+a²c

=>1/a² + 1/b² = 1/c

Thus:--

Your answer will be option number

(1)

Hopes its helps u

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