1.

(a,0) and (b,0) are centres of two circles belonging to a coaxial system of which y-axis is the radical axis. If radius of one of the circle is r then the radius of the other circles is

Answer»

`(r^(2) + B^(2) + a^(2))^(1//2)`
`(r^(2) + b^(2) - a^(2))^(1//2)`
`(r^(2) + b^(2) - a^(2))^(1//3)`
`(r^(2) + b^(2) + a^(2))^(10)`

Solution :Given centres are, `C_(1) = (a, 0) C_(1) = (b, 0)` and `r_(1) = r`
LET the radius of other circle be `r_(2)`
Equation of circles are
`(x - a)^(2) + y^(2) = r^(2)` and `(x - b)^(2) + y^(2) = r_(2)^(2)`
`implies S -= x^(2) + y^(2) 2ax - r^(2) + a^(2) = 0` and `S^(1) -= x^(2) + y^(2) - 2bx - r_(2)^(2) + b^(2) = 0`
Radical AXIS is, `S - S^(1) = 0`
`implies -2ax - r^(2) + 2bx + r_(2)^(2) + a^(2) - b^(2) = 0`
`implies` since radical axis is y-axis `implies x = 0`
`implies r_(2)^(2) = r^(2) + a^(2) - b^(2) = 0`
`implies r_(2)^(2) = r^(2) + b^(2) - a^(2)`
`:. r_(2) = (r^(2) + b^(2) - a^(2)y^(2))`


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