Question

The point $$P$$ moves in the plane of a regular hexagon such that the sum of the squares of its distances from the vertices of the hexagon is $$6{a^2}.$$  If the radius of the circumcircle of the hexagon is $$r\left( { < a} \right)$$  then the locus of $$P$$ is :

A. a pair of straight lines
B. an ellipse
C. a circle of radius $$\sqrt {{a^2} - {r^2}} $$  
D. an ellipse of major axis $$a$$ and minor axis $$r$$
Answer :   a circle of radius $$\sqrt {{a^2} - {r^2}} $$
Solution :
Circle mcq solution image
The vertices are $$A\left( {r,\,0} \right),\,B\left( {r\cos \,{{60}^ \circ },\,r\sin \,{{60}^ \circ }} \right),\,C\left( { - r\cos \,{{60}^ \circ },\,r\sin \,{{60}^ \circ }} \right),\,D\left( { - r,\,0} \right),\,E\left( { - r\cos \,{{60}^ \circ },\, - r\sin \,{{60}^ \circ }} \right),\,F\left( {r\cos \,{{60}^ \circ },\, - r\sin \,{{60}^ \circ }} \right).$$
If $$P = \left( {x,\,y} \right)$$   then $$\sum {P{A^2} = 6{a^2}} $$    implies
$$\eqalign{ & \left\{ {{{\left( {x - r} \right)}^2} + {y^2}} \right\} + \left\{ {{{\left( {x - \frac{r}{2}} \right)}^2} + {{\left( {y - \frac{{r\sqrt 3 }}{2}} \right)}^2}} \right\} + ..... + \left\{ {{{\left( {x - \frac{r}{2}} \right)}^2} + {{\left( {x + \frac{{r\sqrt 3 }}{2}} \right)}^2}} \right\} = 6{a^2} \cr & {\text{or }}2\left( {{x^2} + {y^2} + {r^2}} \right) + 2\left( {{x^2} + {y^2} + \frac{{{r^2}}}{4} + \frac{{3{r^2}}}{4}} \right) + 2\left( {{x^2} + {y^2} + \frac{{{r^2}}}{4} + \frac{{3{r^2}}}{4}} \right) = 6{a^2} \cr & {\text{or }}{x^2} + {y^2} + {r^2} = {a^2} \cr & {\text{or }}{x^2} + {y^2} = {\left( {\sqrt {{a^2} - {r^2}} } \right)^2} \cr} $$

Releted MCQ Question on
Geometry >> Circle

Releted Question 1

A square is inscribed in the circle $${x^2} + {y^2} - 2x + 4y + 3 = 0.$$      Its sides are parallel to the coordinate axes. The one vertex of the square is-

A. $$\left( {1 + \sqrt 2 ,\, - 2 } \right)$$
B. $$\left( {1 - \sqrt 2 ,\, - 2 } \right)$$
C. $$\left( {1 - 2 ,\, + \sqrt 2 } \right)$$
D. none of these
Releted Question 2

Two circles $${x^2} + {y^2} = 6$$    and $${x^2} + {y^2} - 6x + 8 = 0$$     are given. Then the equation of the circle through their points of intersection and the point $$\left( {1,\,1} \right)$$  is-

A. $${x^2} + {y^2} - 6x + 4 = 0$$
B. $${x^2} + {y^2} - 3x + 1 = 0$$
C. $${x^2} + {y^2} - 4y + 2 = 0$$
D. none of these
Releted Question 3

The centre of the circle passing through the point (0, 1) and touching the curve $$y = {x^2}$$   at $$\left( {2,\,4} \right)$$  is-

A. $$\left( {\frac{{ - 16}}{5},\,\frac{{27}}{{10}}} \right)$$
B. $$\left( {\frac{{ - 16}}{7},\,\frac{{53}}{{10}}} \right)$$
C. $$\left( {\frac{{ - 16}}{5},\,\frac{{53}}{{10}}} \right)$$
D. none of these
Releted Question 4

The equation of the circle passing through $$\left( {1,\,1} \right)$$  and the points of intersection of $${x^2} + {y^2} + 13x - 3y = 0$$      and $$2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$$      is-

A. $$4{x^2} + 4{y^2} - 30x - 10y - 25 = 0$$
B. $$4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$$
C. $$4{x^2} + 4{y^2} - 17x - 10y + 25 = 0$$
D. none of these

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