The equation of the circle with radius sqrt{1 | vedclass

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Question

(A) Let the centre of the circle be $(a, 0)$ where $a > 0$ because it lies on the positive $x$-axis.Given the radius $r = \sqrt{17}$,the equation of t

Vedclass · Accepted Answer

$x^{2} + y^{2} - 8x - 1 = 0$

Vedclass · Answer

(A) Let the centre of the circle be $(a, 0)$ where $a > 0$ because it lies on the positive $x$-axis.Given the radius $r = \sqrt{17}$,the equation of the circle is $(x - a)^{2} + (y - 0)^{2} = r^{2}$.Substituting the values,we get $(x - a)^{2} + y^{2} = 17$.Since the circle passes through the point $(0, 1)$,we substitute $x = 0$ and $y = 1$ into the equation:$(0 - a)^{2} + 1^{2} = 17$$a^{2} + 1 = 17$$a^{2} = 16$Since $a > 0$,we have $a = 4$.Substituting $a = 4$ back into the circle equation:$(x - 4)^{2} + y^{2} = 17$$x^{2} - 8x + 16 + y^{2} = 17$$x^{2} + y^{2} - 8x - 1 = 0$.

The equation of the circle with radius $\sqrt{17}$ units,whose centre lies on the positive side of the $x$-axis and which passes through the point $(0, 1)$,is:

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