What the distance between the centre of the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and a point situated outside, $p (x_{1}, y_{1}) .$

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Solution

The correct option is A

Given circle is,

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

$i . e ., (x + g)^{2} + (y + f)^{2} - g^{2} - f^{2} + c = 0$ [ By completion of squares]

$i . e ., (x + g)^{2} + (y + f)^{2} - g^{2} - f^{2} + c$

The standard equation of the circle,

$(x - h)^{2} + (y - k)^{2} = r^{2}$ , Where (h,k) is centre $&$ r is radius

$∴$ Centre and radius of given circle is,

$O \equiv (- g, - h)$ radius $= \sqrt{g^{2} + f^{2} - c}$

Now we have,

External point $e q u i v P (x_{1}, y_{1})$

Centre of circle $\equiv C (- g, - f)$

$∴$ Distance between them by distance formulae.

$d = \sqrt{(x_{1} + g)^{2} + (y_{1} + f)^{2}}$

Hence answer is $(a)$

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What the distance between the centre of the circle x2+y2+2gx+2fy+c=0 and a point situated outside, p(x1,y1).

What the distance between the centre of the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and a point situated outside, $p (x_{1}, y_{1}) .$