If it is possible to draw a triangle which circumscribe the circle${(x - (\alpha - 2\beta ))^2} + {(y - (\alpha + \beta ))^2} = 1$ and is inscribed by${x^2} + {y^2} - 2x - 4y + 1 = 0$, then
1. $\beta = \dfrac{{ - 1}}{3}$
2. $\beta = \dfrac{2}{3}$
3. $\alpha = \dfrac{5}{3}$
4. $\alpha = \dfrac{{ - 5}}{2}$

Question

If it is possible to draw a triangle which circumscribe the circle${(x - (\alpha - 2\beta ))^2} + {(y - (\alpha + \beta ))^2} = 1$ and is inscribed by${x^2} + {y^2} - 2x - 4y + 1 = 0$, then
1. $\beta = \dfrac{{ - 1}}{3}$
2. $\beta = \dfrac{2}{3}$
3. $\alpha = \dfrac{5}{3}$
4. $\alpha = \dfrac{{ - 5}}{2}$

Vedantu Content Team · Accepted Answer

Hint: Focus on showing that the circumcenter and incenter are at the same point by using the formula of distance between incenter and circumcenter i.e. Distance = $\sqrt {{R^2} - 2rR} $and show that the distance between incenter and circumcenter is 0 and find the value of \[\alpha \] and$\beta $,

Complete step-by-step answer:

Let’s construct the rough diagram using the given information

Equation of incircle is ${(x - (\alpha - 2\beta ))^2} + {(y - (\alpha + \beta ))^2} = 1$

And equation of circumcircle is ${x^2} + {y^2} - 2x - 4y + 1 = 0$

\[ \Rightarrow \]${(x - 1)^2} + {(y - 2)^2} = 4$

So the radius of incircle r = 1 and the radius of circumcircle R = 2

Since we have the radius of incircle and circumcircle

Applying the formula of distance between P and O i.e. distance = $\sqrt {{R^2} - 2rR} $

Distance = \[\sqrt {{2^2} - 2 \times 1 \times 2} \]

Distance = 0

Since the distance between P and O is 0

Therefore circumcenter and incenter are at the same point

The general equation of the circle is \[{(x - h)^2} + {(y - k)^2} = {r^2}\]where (h, k) is the center of the circle.

Since both incircle and circumcircle have the same center therefore

By the of equation 1 and 2

$\alpha - 2\beta = 1$ -(Equation 1)

$\alpha + \beta = 2$ -(Equation 2)

Subtracting equation 2 by equation 1

$\alpha + \beta - \alpha + 2\beta = 2 - 1$

$ \Rightarrow $$\beta = \dfrac{1}{3}$

Putting the value of \[\alpha \]in equation 1

$\alpha - 2 \times \dfrac{1}{3} = 1$

Therefore value \[\alpha = \dfrac{5}{3}\].

Note: In these types of questions constructing a rough diagram using the given information in the question using the distance formula to find the distance between P and O. Then show that both incircle and circumcircle have the same center next comparing the equations of both circles and find the value of \[\alpha \]and$\beta $.