Circles are drawn through the point (2,0) to | vedclass

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(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through $(2,0)$,we have $4+4g+c=0$,so $c = -4-4g$. The intercept

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$x^2+y^2-9x-2ky+14=0, k \in R^{+}$

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(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through $(2,0)$,we have $4+4g+c=0$,so $c = -4-4g$. The intercept on the $X$-axis is given by $2\sqrt{g^2-c} = 5$. Squaring both sides,$4(g^2-c) = 25$. Substituting $c = -4-4g$,we get $4(g^2+4g+4) = 25$,which simplifies to $4g^2+16g+16=25$,or $4g^2+16g-9=0$. Solving for $g$,$g = \frac{-16 \pm \sqrt{256-4(4)(-9)}}{8} = \frac{-16 \pm \sqrt{256+144}}{8} = \frac{-16 \pm 20}{8}$. Thus,$g = \frac{4}{8} = 0.5$ or $g = \frac{-36}{8} = -4.5$. Since the centre $(-g, -f)$ lies in the first quadrant,$-g > 0$,so $g Then $c = -4-4(-4.5) = -4+18 = 14$. The equation is $x^2+y^2-9x+2fy+14=0$. Let $2f = -2k$,where $k \in R^{+}$,so the equation is $x^2+y^2-9x-2ky+14=0$.

Circles are drawn through the point $(2,0)$ to cut intercepts of length $5$ units on the $X$-axis. If their centre lies in the first quadrant,then their equation is

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