Question

x^+y^-4x-2y-20==0 then the diameter of the circle is

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ diameter \: of \: circle \\ \\ so \: for \: a \: certain \: circle \\ given \: equation \: is \: \\ x {}^{2} + y {}^{2} - 4x - 2y - 20 = 0 \\ \\ comparing \: it \: with \\ x {}^{2} + y {}^{2} + 2gx + 2fy + c = 0 \\ \\ we \: have \: then \\ 2g = - 4 \\ g = - 2 \\ \\ 2f = - 2 \\ f = - 1 \\ \\ c = - 20$

$so \: we \: know \: that \\ \\ radius \: of \: circle \: (r) = \sqrt{g {}^{2} + f {}^{2} - c} \\ \\ = \sqrt{( - 2) {}^{2} + ( - 1) {}^{2} - ( - 20)} \\ \\ = \sqrt{4 + 1 + 20} \\ \\ = \sqrt{25} \\ \\ r= 5 \\ \\ thus \: then \: \\ diameter \: (d) = 2r \\ \\ = 2 \times 5 \\ \\ d= 10.units$