Three distinct points  A,B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( –1, 0) is equal to 1/3. Then the circumcentre of the triangle ABC is at the point

  • Option 1)

    \left ( \frac{5}{4},0 \right )

  • Option 2)

    \left ( \frac{5}{2},0 \right )

  • Option 3)

    \left ( \frac{5}{3},0 \right )

  • Option 4)

    \left ( 0,0 \right )

 

Answers (1)

As we learnt in 

Circumcentre -

The point of intersection of the perpendicular bisectors of the sides of a triangle.

- wherein

The centre of the circumcircle of a triangle.

 

 Let x be (h,k)

A,B, C lie on locus of x.

given that \frac{PA}{PB}=\frac{1}{3}

\Rightarrow 9PA^{2}=9PB^{2}

\Rightarrow 9\left(\sqrt{(h^{2}-1^{2})+k^{2}} \right )^{2}=(h+1)^{2}+k^{2}

\Rightarrow 8h^{2}+8k^{2}-20h+8=0

\Rightarrow h^{2}+k^{2}-\frac{5}{2}h+1=0

Centre is (\frac{5}{4},0)

Circumstance is (\frac{5}{4},0)

 


Option 1)

\left ( \frac{5}{4},0 \right )

This option is correct.

Option 2)

\left ( \frac{5}{2},0 \right )

This option is incorrect.

Option 3)

\left ( \frac{5}{3},0 \right )

This option is incorrect.

Option 4)

\left ( 0,0 \right )

This option is incorrect.

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