# NCERT Exemplar Solutions Class 10 Mathematics Solutions for Coordinated Geometry - Exercise 7.1 in Chapter 7 - Coordinated Geometry

Question 12 Coordinated Geometry - Exercise 7.1

A circle drawn with origin as the centre passes through a point (13/2,0). The point which does not lie in the interior of the circle is

Radius of circle = \sqrt{\left(\frac{13}{2}-0\right)^{2}+(0-0)^{2}}=\frac{13}{2}=6.5 \text { units }

(a) \ Distance \ point \ \left(\frac{-3}{4}, 1\right) \ from \ (0,0) \ is \\ =\sqrt{\left(\frac{-3}{4}-0\right)^{2}+(1-0)^{2}}=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}=1.25 \text{ units} \\ The \ distance \ 1.25<6.5 \ . \ SO\ , \ the \ point \left(\frac{-3}{4}, 1\right) lies \ in \ the \ interior \ of \ the \ circle.

(b) Distance of point (2, 7/3) from (0, 0) is

=\sqrt{(2-0)^{2}+\left(\frac{7}{3}-0\right)^{2}}=\sqrt{4+\frac{49}{9}}=\sqrt{\frac{85}{9}}=\frac{9.2195}{3}=3.0731<6.25\\ \text { So, the point }\left(2, \frac{7}{3}\right) \text { lies in the interior of the circle. }

(c) Distance of point (5, -1/2) from (0, 0) is

=\sqrt{(5-0)^{2}+\left(-\frac{1}{2}-0\right)}=\sqrt{25+\frac{1}{4}}=\sqrt{\frac{101}{4}}=\frac{10.0498}{2}=5.0249<6.5\\ \text { So, the point }\left(5, \frac{-1}{2}\right) \text { lies in the interior of the circle. }

(d) Distance of point (-6, 5/2) from (0, 0) is

=\sqrt{(-6-0)^{2}+\left(\frac{5}{2}-0\right)^{2}}=\sqrt{36+\frac{25}{4}}=\sqrt{\frac{169}{4}}=\frac{13}{2}=6.5 \text { units }\\ \text { So, }\left(-6, \frac{5}{2}\right) \text { lies on the circle. It does not lie in the interior of the circle. }

Hence, (d) is the correct answer

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