Question

The centre of circle through origin, ( p , o ) and (o , q) is?

Answer 1

Answer:

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Answer 2

The centre of the circle is at $(\dfrac{p}{2},\dfrac{q}{2})$

Tips:

We know that, the general equation of a circle is given b

$x^{2}+y^{2}+2gx+2fy+c=0$

and its centre is at $(-g,-f)$

Step-by-step explanation:

Let, the equation of the circle be

$x^{2}+y^{2}+2gx+2fy+c=0$ ... ... (1)

Since circle (1) passes through the origin (0, 0), (p, 0) and (0, q), we have

• $c=0$

• $p^{2}+2gp=0\Rightarrow g=-\dfrac{p}{2}$

• $q^{2}+2fq=0\Rightarrow f=-\dfrac{q}{2}$

So, we have: $-g=\dfrac{q}{2}$ and $-f=\dfrac{q}{2}$

Thus the centre of the circle is at $(\dfrac{p}{2},\dfrac{q}{2})$

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