Question

Number of the points on the circle $2{{x}^{2}}+2{{y}^{2}}-3x=0$which are at the distance 2 from the point (-2,1) is
a)2
b)0
c)1
d)None of these

Answer 1

Hint: First consider the equation of the given circle then compare it to find the centre of the circle. Then find distance between the given point and centre, after that compare it with radius to get what is asked.

Complete step-by-step answer:

We see that in the above given question the equation of circle is, $2{{x}^{2}}+2{{y}^{2}}-3x=0$

Which can also be written as

${{x}^{2}}+{{y}^{2}}-\left( \dfrac{3}{2} \right)x=0.......(i)$

From the equation (i) of circle,

We can find the centre of circle by formula $(-g,-f)$ and radius of circle $\sqrt{({{g}^{2}}-{{f}^{2}}-c)}$

So, by the formula the Center C of the circle is $\left( \dfrac{3}{4},0 \right)$and r the radius of circle is $\dfrac{3}{4}$

So, now let consider the given point (−2,1) be P

So,

 $CP=\sqrt{\left( {{\left( \dfrac{3}{4}+2 \right)}^{2}}+{{1}^{2}} \right)}$

$=\sqrt{\dfrac{(137)}{2}>\dfrac{11}{2}}$

And \[CP-2>\dfrac{11}{2}-2=\dfrac{7}{2}>r\]

The point P lies outside the given circle.

Hence there is no point on the circle at a distance 2 unit from point P on the given circle.


Note: There might be a confusion in the students for the location of point P. So please check the diagram carefully.

Hence option (b) is the correct answer.