Question

Two circles touch externally. The sum of their areas is 73pi sq. cm and the distance between their centres is 11 cm. Find the radii of the circles​

Answer 1

Answer:

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Step-by-step explanation:

$so \: let \: the \: radii \: of \: two \:externally \\ touching \: circles \: \: be \: \\ x \: and \: y \: respectively \\where \: x > y \\ so \: we \: know \: that \: \\ area \: of \: circle \: = \pi.r {}^{2} \\ \\ therefore \: according \: to \: 1st \: case \\ \pi.x {}^{2} + \pi.y {}^{2} = 73\pi \\ \pi(x{}^{2} + y {}^{2} ) = 73\pi \\ x {}^{2} + y {}^{2} = 73 \\ \\ we \: know \: that \\ whenever \: two \: circles \: touch \: each \: other \: \\ externally \: at \: point \: of \: contact \\ the \: distance \: \: between \: their \: centres \: \\ is \: the \: sum \: of \: their \: respective \: radii \\ \\ according \: to \: second \: condition \\ x + y = 11 \\ x = 11 - y \: \: \: \: \: \: \: \: (1)$

$x {}^{2} + y {}^{2} = 73 \\ \\ (11 - y) {}^{2} + y {}^{2} = 73 \\ \\ 121 + y {}^{2} + y {}^{2} - 22y = 73 \\ \\ 2y {}^{2} - 22y + 48 = 0 \\ \\ y {}^{2} - 11y + 24 = 0 \\ \\ y {}^{2} - 8y - 3y + 24 = 0 \\ \\ y(y - 8) - 3(y - 8) = 0 \\ \\ (y - 8)(y - 3) = 0 \\ \\ y = 8 \: \: \: or \: \: \: y = 3$

$but \: if \: we \: take \: \\ y = 8 \\then \: \: x = 3 \\ and \: as \: here \: \: x > y \\ y = 8 \: is \: absurd \\ hence \: then \\ \\ y = 3.cm \\ \\ therefore \: accordingly \\ x = 8.cm$