Question

A circle passes through the origin and the point(5/2,1/2) and has 2y-3x=0 as a diameter. Find its equation.​

Answer 1

$\text{Let the equation of circle be}$

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

$\text{since it passes through (0,0) and (\frac{5}{2},\frac{1}{2}) , we have}$

$0^2+0^2+2(0)x+2(0)y+c=0$

$\implies\boxed{\bf\,c=0}$

$\text{and}$

$\frac{25}{4}+\frac{1}{4}+2g(\frac{5}{2})+2f(\frac{1}{2})+0=0$

$\frac{13}{2}+5g+f=0$

$\implies\,13+10g+2f=0$

$\implies\,10g+2f=-13$ ........(1)

$\text{Since the centre (-g,-f) lies on 2y-3x,}$

$3g-2f=0$........(2)

$\text{Adding (1) and (2), we get}$

$13g=-13$

$\implies\boxed{\bf\,g=-1}$

$(2)\implies\;-3=2f$

$\implies\boxed{\bf\,f=\frac{-3}{2}}$

$\therefore\text{The equation of the required circle is}$

$x^2+y^2-2x-3y+0=0$

$\implies\boxed{\bf\,x^2+y^2-2x-3y=0}$

Find more:

Find the equation of circle which passes through points (5,-8),(2,-9) and(2,1). Find also the coordinates of its centre and radius

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Answer 2

Step-by-step explanation:

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