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  • A circle drawn with origin as the Centre passes through (13/2,0) . The point which does not lie in the interior of the circle is : (A) (B) (C) (D)

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

\\(A) \frac{-3}{4}, 1 \\(B) 2, \frac{7}{3} \\(C) 5, \frac{-1}{2} \\(D) \left(-6, \frac{5}{2}\right)

Answers (1)

A) Distance of the point (-3/4, 1) 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 \mathrm{units}

The distance  1.25<6.5 .  so the point (-3/4, 1)  is lie
B) Distance point (2, 7/3) from (0,0) is

=\sqrt{\left(\frac{7}{3}-0\right)^{2}+(2-0)^{2}}=\sqrt{\frac{49}{9}+4}=\sqrt{\frac{85}{9}}=\frac{9.2195}{3}=3.0731<6.25

So the point is lie
C) Distance point (5,-1/2) from (0,0) is

\\=\sqrt{\left(-\frac{1}{2}-0\right)^{2}+(5-0)^{2}}=\sqrt{\frac{1}{4}+25}=\sqrt{\frac{101}{4}}=\frac{10.0498}{2}=5.0249<6.25

So the point is lie in the interior of the circle
D)
The circle passes through (13/2, 0) having a centre (0,0)

\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(13 / 2,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,0)

\text{Apply distance formula}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}

\text{Radius}=\sqrt{(6.5-0)^{2}+(0-0)^{2}}
\\=\sqrt{(6.5)^{2}}=6.5$ \\$=5^{2}+0.5^{2}-6.5^{2} \quad (Negative) \\\left(-6, \frac{5}{2}\right)=(-6)^{2}+(2.5)^{2}-6.5$ \\$=6^{2}+2.5^{2}-6.5 \quad(positive)

Hence, point D is the correct answers.

Posted by

infoexpert21

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