Question

tangets ab and ac are drawn to the circle x2+y2-2x+4y+1=0 from a(0,1) then equation of circle passing through a,b and c is

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ equation \: of \: circle \: passing \: through \\ A,B,C \\ (equation \: of \: circumcir cle \: of \: △ABC) \\ \\ so \: for \: a \: circle \: with \: centre \: X \\ its \: equation \: is \\ x {}^{2} + y {}^{2} - 2x + 4y + 1 = 0 \\ \\ comparing \: it \: \: with \\ x {}^{2} + y {}^{2} + 2gx + 2fy + c = 0 \\ we \: get \\ \\ 2g = - 2 \\ g = - 1 \\ \\ 2f = 4 \\ f = 2 \\ \\ (h,k) = ( - g, - f) = (1, - 2) \\ \\ so \: since \: circle \: with \: centre \: Y \\ passes \: through \: O \: and \: A \\ O \: and \: A \: are \: the \: end - points \\ of \: its \: diameter \\ and \: it \: circumscribes \: △ABC$

$so \: thus \: then \\ O = (x1,y1) = (1, - 2) \\ A = (x2,y2) = (0,1) \\ \\ we \: know \: that \\ equation \: of \: any \: circle \: wih \: end \: \\ points \: of \: diameter \: is \: given \: by \\( x - x1)(x - x2) + (y - y1)(y1 - y2) = 0 \\ \\ (x - 1)(x - 0) + (y - ( - 2))(y - 1) = 0 \\ \\ x(x - 1) + (y + 2)(y - 1) = 0 \\ \\ x {}^{2} - x + y {}^{2} - y + 2y - 2 = 0 \\ \\ x {}^{2} + y {}^{2} - x + y - 2 = 0 \\ is \: our \: required \: equation \: of \: circle$