Question

find the equation of a circle passing through the points (0, 0) (a,0) and (0, b)​

Answer 1

Step-by-step explanation:

We have circle through the point A(0, 0), B(a, 0) and C(0, b).

Clearly triangle is right angled at vertex A.

So, centre of the circle is the mid point of hypotenuse BC which is (a/2, b/2).

Answer 2

The equation of circle is $x^2+y^2-ax-by=0$

Step-by-step explanation:

The passing points of circle are (0, 0) (a,0) and (0, b)​

Let the general equation of circle:

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

• For point: (0,0)

$0^2+0^2+2g\times 0+2f\times 0+c=0$

$c=0$

• For point: (a,0) and c=0

$a^2+0^2+2ga+2f\times 0+0=0$

$g=-\dfrac{a}{2}$

• For point: (0,b) and c=0

$0^2+b^2+2g\times 0+2fb+0=0$

$f=-\dfrac{b}{2}$

Substitute g, f and c into formula

Equation of circle:

$x^2+y^2-ax-by=0$

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Equation of circle

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