show that the equation x^2 +y^2-2x-2ay-8=0 represents for different values of a ,a system of circle passing through two fixed points a b on the x axis and find the equation of that circle of the system  the tangents to which at A,B meet on the line x+2y+5=0

Dear student

The given equation can be written as

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{ay}-8=0 \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2-2\mathrm{ay}+{\mathrm{a}}^2=1+{\mathrm{a}}^2+8 \\ \Rightarrow {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-\mathrm{a}\right)}^2={\mathrm{a}}^2+9 \end{gathered}$$

Hence this represents a system of circles, for different values of a, with centre (1,a) and radius of $\sqrt{a^2+9}$
This circle will pass through two points on x-axis A and B, which we can get by putting y =0

$$\begin{gathered} {\left(\mathrm{x}-1\right)}^2+{\left(0-\mathrm{a}\right)}^2={\mathrm{a}}^2+9 \\ \Rightarrow {\left(\mathrm{x}-1\right)}^2+{\mathrm{a}}^2={\mathrm{a}}^2+9 \\ \Rightarrow \mathrm{x}-1=\pm 3 \\ \Rightarrow \mathrm{x}=-2,4 \end{gathered}$$

So the two fixed points are A and B as (4,0) and (-2,0)

The slope of the tangent of the circle is given by differentiating the eqn

$$\begin{gathered} 2\left(\mathrm{x}-1\right)+2\left(\mathrm{y}-\mathrm{a}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}-1}{\mathrm{a}-\mathrm{y}} \\ \mathrm{Now}, {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(4,0\right)} = \frac{3}{\mathrm{a}} \\ \mathrm{Now}, {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(-2,0\right)} = -\frac{3}{\mathrm{a}} \end{gathered}$$

So the equation of tangents to circle at A and B - (4,0) and (-2,0) are

$$\begin{gathered} \mathrm{y}-0=\frac{3}{\mathrm{a}}\left(\mathrm{x}-4\right) \\ \mathrm{y}-0=\frac{-3}{\mathrm{a}}\left(\mathrm{x}+2\right) \end{gathered}$$

The above two lines intersect at (1,-9/a)

As the tangents meet on the line x+2y+5=0

$$\begin{gathered} 1+2\times \frac{-9}{\mathrm{a}}+5=0 \\ \Rightarrow \frac{18}{\mathrm{a}}=6 \\ \Rightarrow \mathrm{a}=3 \\ \mathrm{Hence} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{becomes} \\ {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-3\right)}^2=18 \end{gathered}$$

Regards