Question

Find the equation of the circle with passing through three points $(0,2),(3,0)$ and $(3,2)$.

• A. $x^2+y^2-3x-2y=0$
• B. $x^2+y^2+3x-2y=0$
• C. $x^2+y^2-3x+2y=0$
• D. $x^2+y^2+3x+2y=0$

Answer 1

Answer: (A)

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

Three points on the circle are$:-A\equiv (0,2),B\equiv (3,0),C\equiv (3,2)$

Let $P\equiv (x,y)$ be its centre, then $|PA|=|PB|=|PC|$

$\Rightarrow \sqrt{(x-0{)}^2+(y-2{)}^2}=\sqrt{(x-3{)}^2+(y-0{)}^2}=\sqrt{(x-3{)}^2+(y-2{)}^2}$

$\Rightarrow \sqrt{x^2+y^2+4-4y}=\sqrt{x^2+9-6x+y^2}=\sqrt{x^2+9-6x+y^2+4-4y}$

$\Rightarrow 4-4y=9-6x=9-6x+4-4y$

Taking first and third parts of the above equation,

$4-4y=9+4-6x-4y\Rightarrow 6x=9\Rightarrow x=\frac{9}{6}=\frac{3}{2}$

Taking second and third parts of the above equation,

$9-6x=9+4-6x-4y\Rightarrow 4y=4\Rightarrow y=1\therefore P\equiv (\frac{3}{2},1)$

So, radius $r=PA=\sqrt{(\frac{3}{2}{)}^2+(1-2{)}^2}=\sqrt{\frac{9}{4}+1}=\frac{\sqrt{5}}{2}$

Thus, equation of the circle ;_$(x-\frac{3}{2}{)}^2+(y-1{)}^2=(\frac{\sqrt{5}}{2}{)}^2$

$\Rightarrow x^2+y^2-3x-2y+1-\frac{5}{4}+\frac{9}{4}=0\Rightarrow x^2+y^2-3x-2y=0$