A circle passes through the points (-1, 1),(0 | vedclass

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Substituting the points $(-1, 1)$,$(0, 6)$,and $(5, 5)$:$1 + 1 - 2g + 2f + c =

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$(B)$ or $(C)$ both

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Substituting the points $(-1, 1)$,$(0, 6)$,and $(5, 5)$:$1 + 1 - 2g + 2f + c = 0 \implies -2g + 2f + c = -2$ $(i)$$36 + 12f + c = 0 \implies 12f + c = -36$ (ii)$25 + 25 + 10g + 10f + c = 0 \implies 10g + 10f + c = -50$ (iii)Solving these equations,we get $g = -2$,$f = -3$,and $c = 0$. The circle is $x^2 + y^2 - 4x - 6y = 0$.The centre is $(2, 3)$ and radius $r = \sqrt{2^2 + 3^2} = \sqrt{13}$.The line joining the origin $(0, 0)$ to the centre $(2, 3)$ has slope $m = \frac{3-0}{2-0} = \frac{3}{2}$.The tangent at a point $(x_1, y_1)$ is parallel to this line if its slope is $\frac{3}{2}$.The slope of the tangent at $(x_1, y_1)$ is $-\frac{x_1 - 2}{y_1 - 3} = \frac{3}{2} \implies 2x_1 - 4 = -3y_1 + 9 \implies 2x_1 + 3y_1 = 13$.Substituting $x_1 = \frac{13 - 3y_1}{2}$ into the circle equation $(x_1 - 2)^2 + (y_1 - 3)^2 = 13$:$(\frac{13 - 3y_1 - 4}{2})^2 + (y_1 - 3)^2 = 13 \implies (\frac{9 - 3y_1}{2})^2 + (y_1 - 3)^2 = 13 \implies \frac{9}{4}(y_1 - 3)^2 + (y_1 - 3)^2 = 13 \implies \frac{13}{4}(y_1 - 3)^2 = 13 \implies (y_1 - 3)^2 = 4$.So $y_1 - 3 = \pm 2$,giving $y_1 = 5$ or $y_1 = 1$.If $y_1 = 5$,$x_1 = \frac{13 - 15}{2} = -1$. Point is $(-1, 5)$.If $y_1 = 1$,$x_1 = \frac{13 - 3}{2} = 5$. Point is $(5, 1)$.

$A$ circle passes through the points $(-1, 1)$,$(0, 6)$,and $(5, 5)$. The point$(s)$ on this circle,the tangent$(s)$ at which is/are parallel to the straight line joining the origin to its centre is/are:

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