Question

Let \[{C_{1,}}{C_2}\] be two circles touching each other externally at the point A and let AB be the diameter of the circle ${C_1}$. Draw a secant $B{A_3}$ to circle ${C_2}$, intersecting to circle ${C_1}$ at a point ${A_1}$($ \ne A)$ , and circle ${C_2}$ at points ${A_2}$ and ${A_3}$ .If $B{A_1}$=2, $B{A_2} = 3$ and $B{A_3} = 4$ ,then the radii of the circles ${C_1}$ and ${C_2}$ are respectively.

Answer 1

Hint: On drawing the figure with the data given, we get two triangles, let us prove that those triangles are similar and solve the problem further.

Complete step-by-step answer:

On drawing the figure with respect to the given data, we get two triangles here,
That is $\vartriangle B{C_1}E$ and $\vartriangle B{C_2}F$
Now ,let us compare these two triangles
So, in $\vartriangle B{C_1}E$ and $\vartriangle B{C_2}F$,we can compare and write
$
  \angle {C_1}EB = \angle {C_2}FB = {90^ \circ } \\
  \angle {C_1}BE = \angle {C_2}BF = {\text{common angle}} \\
  \angle B{C_1}E = \angle B{C_2}F = {\text{remaining angle}} \\
    \\
$
So, therefore we can write $\vartriangle B{C_1}E$ and $\vartriangle B{C_2}F$ are similar to each other(from AA postulate)

So , from this we can write
$\dfrac{{B{C_1}}}{{B{C_2}}} = \dfrac{{BE}}{{BF}}$ (Corresponding sides of two similar triangles are proportional)
On comparing with the figure, we can substitute the value of $B{C_1}$ ,$B{C_2}$ , BE and BF as follows
$\dfrac{{{r_1}}}{{{r_1} + {r_1} + {r_2}}} = \dfrac{1}{{3 + \dfrac{1}{2}}}$ (Since, centre ${C_1}$ and ${C_2}$ bisects $B{A_{1,}}{A_2},{A_3}{\text{ with }} \bot {\text{ at E,F}}$ )
On simplifying this further we get $3{r_1} = 2{r_2}$
$ \Rightarrow {r_2} = \dfrac{3}{2}{r_1}$
Also, if we consider P as an external point from which a secant is cutting a circle at point A,B then we can write $PA \times PB$ =constant
So, we can write $B{C_1} \times B{C_2} = BE \times BF$
Let us substitute the values of $B{C_1},B{C_2}$ ,BE ,BF in the equation, so we get
$
   \Rightarrow 4{r_1}({r_1} + {r_2}) = 4 \times 3 \\
   \Rightarrow {r_1}^2 + \dfrac{3}{2}{r_1}^2 = 3 \\
   \Rightarrow {r_1}^2 = \dfrac{6}{5} \\
   \Rightarrow {r_1} = \dfrac{{\sqrt {30} }}{5} \\
   \Rightarrow {r_{2 = }}\dfrac{3}{2}\dfrac{{\sqrt {30} }}{5} = \dfrac{{3\sqrt {30} }}{{10}} \\
$
 So , we get the radius of circle ${C_1}$ =$\dfrac{{\sqrt {30} }}{5}$ and radius of circle ${C_2} = \dfrac{{3\sqrt {30} }}{{10}}$

Note: The secant of a circle is the line that intersects the circle at minimum two points, also if a perpendicular is drawn from the centre of the circle to the secant then it bisects it. Take the data properly from the diagram.