A circle drawn with the origin as the center | vedclass

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(C) The center of the circle is $(0,0)$ and it passes through the point $\left(\frac{13}{2}, 0\right)$.Therefore,the radius of the circle $r$ is the d

Vedclass · Accepted Answer

$\left(-6, \frac{5}{2}\right)$

Vedclass · Answer

(C) The center of the circle is $(0,0)$ and it passes through the point $\left(\frac{13}{2}, 0\right)$.Therefore,the radius of the circle $r$ is the distance between $(0,0)$ and $\left(\frac{13}{2}, 0\right)$.$r = \sqrt{\left(\frac{13}{2}-0\right)^{2} + (0-0)^{2}} = \sqrt{\left(\frac{13}{2}\right)^{2}} = \frac{13}{2} = 6.5$.$A$ point $(x, y)$ lies in the interior of the circle if its distance from the center is less than $r$,on the circle if the distance is equal to $r$,and outside if the distance is greater than $r$.$(a)$ Distance of $\left(-\frac{3}{4}, 1\right)$ from $(0,0) = \sqrt{\left(-\frac{3}{4}\right)^{2} + 1^{2}} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = 1.25 $(b)$ Distance of $\left(2, \frac{7}{3}\right)$ from $(0,0) = \sqrt{2^{2} + \left(\frac{7}{3}\right)^{2}} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{85}{9}} \approx 3.07 $(c)$ Distance of $\left(-6, \frac{5}{2}\right)$ from $(0,0) = \sqrt{(-6)^{2} + \left(\frac{5}{2}\right)^{2}} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144+25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5$. (On the circle)$(d)$ Distance of $\left(5, -\frac{1}{2}\right)$ from $(0,0) = \sqrt{5^{2} + \left(-\frac{1}{2}\right)^{2}} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} \approx 5.02 Thus,the point $\left(-6, \frac{5}{2}\right)$ does not lie in the interior of the circle.

$A$ circle drawn with the origin as the center passes through $\left(\frac{13}{2}, 0\right)$. The point which does not lie in the interior of the circle is

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