Question

Let the tangent drawn at $(-1,2)$ to the circle $x^2+y^2-3x-3y-2=0$ is normal to the circle $x^2+y^2-2ay+b=0.$ If the radius $\left(r\right)$ of the second circle is such that $\left[r\right]=1,$ then
([.] denotes the greatest integer function)

• A. $b\in [48,49)$
• B. $b\in (45,48]$
• C. $b\in (48,49]$
• D. $b\in [45,48)$

Answer 1

The correct option is B $b\in (45,48]$

Given circle is $x^2+y^2-3x-3y-2=0$
Equation of tangent at $(-1,2)$ is
$x{x}_1+y{y}_1-\frac{3}{2}(x+x_1)-\frac{3}{2}(y+y_1)-2=0\Rightarrow -x+2y-\frac{3}{2}(x-1)-\frac{3}{2}(y+2)-2=0\Rightarrow 5x-y+7=0$

We know that this is equation of normal to the circle $x^2+y^2-2ay+b=0$, so it passes through the centre $(0,a)$ of the circle, we get
$0-a+7=0\Rightarrow a=7$

Therefore, the equation of second circle is
$x^2+y^2-14y+b=0$
So, the radius of the circle is
$r=\sqrt{49-b}$
For the squareroot to be defined
$49-b>0\Rightarrow b<49\cdots \left(1\right)$
Given, $\left[r\right]=1$
$\Rightarrow 1\le r<2$
$\Rightarrow 1\le \sqrt{49-b}<2$
$\Rightarrow 1\le 49-b<4$
$\Rightarrow -48\le -b<-45$
$\Rightarrow 45<b\le 48$
$\therefore b\in (45,48]$