Question

if the distance of a point P from (6,0) is twice its distance from the point (1,3) show that the locus of P is a circle find its centre and radius​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: show \: that : \\ locus \: of \: P \: is \: a \: circle \\ to \: find : \\ radius \: of \: circle \\ \\ let \: here \\ A = (6,0) \\ B = (1,3) \\ P = (x,y) \\ \\ since \: here \\ AP=2BP \\ AP {}^{2} =4BP {}^{2} \\ \\ (x - 6) {}^{2} + (y - 0) {}^{2} = \\ 4[(x - 1) {}^{2} + (y - 3) {}^{2} ] \\ \\ x {}^{2} - 12x + 36 + y {}^{2} = \\ 4(x { }^{2} - 2x + 1 + y {}^{2} - 6y + 9) \\ \\ x {}^{2} + y {}^{2} - 12x + 36 = 4x {}^{2} - 8x + \\ 4y {}^{2} - 24y + 40 \\ \\ 3x {}^{2} + 3y {}^{2} + 4x - 24y + 4 = 0 \\ \\ dividing \: throughout \: by \: 3 \\ we \: get \\ \\ x {}^{2} + y {}^{2} + \frac{4}{3} x - 8y + \frac{4}{3} = 0 \\ \\ which \: is \: i n\: form \\ x {}^{2} + y {}^{2} + 2gx + 2fy +c = 0 \\ ie \: in \: general \: form \: of \: circle$

$so \: thus \: then \\ 2g = \frac{4}{3} \: ; \: 2f = - 8 \: ; \: c = \frac{4}{3} \\ \\ g = \frac{2}{3} \: ; \: f = - 4 \\ \\ we \: know \: that \\ r = \sqrt{g {}^{2} + f {}^{2} - c} \\ \\ = \sqrt{( \frac{2}{3}) {}^{2} + ( - 4) {}^{2} - \frac{4}{3} } \\ \\ = \sqrt{ \frac{4}{9} + 16 - \frac{4}{3} } \\ \\ = \sqrt{ \frac{4}{9} - \frac{12}{9} + 16} \\ \\ = \sqrt{ \frac{ - 8}{9} + 16 } \\ \\ = \sqrt{ \frac{(16 \times 9) - 8}{9} } \\ \\ = \sqrt{ \frac{144 - 8}{9} } \\ \\ = \sqrt{ \frac{136}{9} } \\ \\ = \frac{2. \sqrt{34} }{3} \\ \\ which \: is \: > 0 \\ ie \: \: \frac{2. \sqrt{34} }{3} \: > 0 \\ \\ hence \: the \: given \: locus \: is \: a \: circle$