Question

Find the equation of the circle that passes through the points $(0,6),(0,0)$ and $(8,0)$

• A. ${(x-4)}^2+{(y-3)}^2=25$
• B. $(x+4{)}^2+(y+3{)}^2=25$
• C. $(x-3{)}^2+(y-4{)}^2=25$
• D. $(x-4{)}^2+(y-3{)}^2=36$

Answer 1

The correct option is A ${(x-4)}^2+{(y-3)}^2=25$

Let the equation of the general form of the required circle be

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

According to the problem, the above equation of the circle passes through the points $(0,6),(0,0)$ and $(8,0)$. Therefore,

$36+12f+c=0$ ………. (2)

$c=0$ ……………. (3)

$64+16g+c=0$ ……………. (4)

Putting $c=0$ in (2), we obtain $f=-3$. Similarly put $c=0$ in (4), we obtain $g=-4$

Substituting the values of $g,f$ and $c$ in (1), we obtain the equation of the required circle as:

$x^2+y^2+2(-4)x+2(-2)y+0=0$ that is

$x^2+y^2-8x-4y+0=0$ can be rewritten as

$x^2+y^2-8x-4y+16+9=0+16+9$

$(x-4{)}^2+(y-3{)}^2=25$

Therefore, the equation of circle is $(x-4{)}^2+(y-3{)}^2=25$.