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Vector Component

A variable li...

Question

A variable line cuts $x$ -axis at $A$ , $y$ -axis at $B$ where $O A = a, O B = b$ ( $O$ as origin) such that $a^{2} + b^{2} = 1$ then the locus of circumcentre of $Δ O A B$ is -

x^{2} + y^{2} = 4

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x^{2} + y^{2} = \frac{1}{4}

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x^{2} - y^{2} = 4

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x^{2} - y^{2} = \frac{1}{4}

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Solution

The correct option is B $x^{2} + y^{2} = \frac{1}{4}$
The $△ O A B$ is a right angled triangle, right angled at $O$ .
We know that every inscribed angle that subtends a diameter is a right angle.
In the above case the circle circumscribing $△ O A B$ will have $A B$ as its diameter.
If $A = (h, 0)$ and $B = (0, k)$
Then the circumcentre is $C = \frac{h}{2}, \frac{k}{2} = (x, y)$ .
Therefore $2 x = h$ and $2 y = k$
Now $O A = | h | = a$ and $O B = | k | = b$
Since $a^{2} + b^{2} = 1$
Hence $h^{2} + k^{2} = 1$
$(2 x)^{2} + (2 y)^{2} = 1$
Or $4 (x^{2} + y^{2}) = 1$
Or $x^{2} + y^{2} = \frac{1}{4}$

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A variable line cuts x-axis at A, y-axis at B where OA=a,OB=b (O as origin) such that a2+b2=1 then the locus of circumcentre of ΔOAB is -

A variable line cuts $x$ -axis at $A$ , $y$ -axis at $B$ where $O A = a, O B = b$ ( $O$ as origin) such that $a^{2} + b^{2} = 1$ then the locus of circumcentre of $Δ O A B$ is -