Question

The coordinate of the centre of his concentric circles are the roots of the quadratic equation $x^2-5x+6=0$, if one circle passing through the origin and the other passes through $(2,3)$, then the equation are

• A. $x^2+y^2-4x-6y=0and{x}^2+y^2-4x-6y+13=0$
• B. $x^2+y^2-6x-4y=0and{x}^2+y^2-6x-4y+11=0$
• C. Both $\left(A\right)$ and $\left(B\right)$
• D. none of these

Answer 1

Answer: (C)

Both $\left(A\right)$ and $\left(B\right)$

$X^2-5x+6=0(x-3)(x-2)=0\Rightarrow x=2,3(h,k)(2,3)(3,2)$

Let equation concentric circles

$(x-2{)}^2+(y-3{)}^2=r^2(x-2{)}^2+(y-3{)}^2={r_1}^2$

And,

$(x-3{)}^2+(y-2{)}^2=r^2(x-3{)}^2+(y-2{)}^2={r_1}^2$

If $(x-2{)}^2+(y-3{)}^2=r^2$ and $(x-3{)}^2+(y-2{)}^2=r^2$ passes through origin, then

$2^2+3^2=r^{2}r=\sqrt{13}$ and $3^2+2^2=r^{2}r=\sqrt{13}$

If $(x-2{)}^2+(y-3{)}^2={r_1}^2$ and $(x-3{)}^2+(y-2{)}^2={r_1}^2$ passes through $(2,3)$, then

$(2-2{)}^2+(3-3{)}^2={r_1}^2{r}_1=0$ $(2-3{)}^2+(3-2{)}^2={r_1}^2{r}_1=\sqrt{2}$

Equation of circles

$\left(1\right)$ $(x-2{)}^2+(y-3{)}^2=13x^2+y^2-4x-6y=0$ and $(x-3{)}^2+(y-2{)}^2=13x^2+y^2-6x-4y=0$

and

$\left(2\right)$ $(x-2{)}^2+(y-3{)}^2=0x^2+y^2-4x-6y+13=0$ and $(x-3{)}^2+(y-2{)}^2=2x^2+y^2-6x-4y+11=0$