Question
The equation of the circle which touches the axes at a distance $$5$$ from the origin is $${y^2} + {x^2} - 2\alpha x - 2\alpha y + {\alpha ^2} = 0.$$ What is the value of $$\alpha \,?$$
A.
$$4$$
B.
$$5$$
C.
$$6$$
D.
$$7$$
Answer :
$$5$$
Solution :
Coordinates of the centre of given circle $$ = \left( {\alpha ,\,\alpha } \right)$$
and radius $$ = \sqrt {{{\left( \alpha \right)}^2} + {{\left( \alpha \right)}^2} - {\alpha ^2}} = \sqrt {{\alpha ^2}} = \alpha $$
$$\eqalign{ & \therefore \,{\left( {\alpha - 5} \right)^2} + {\left( \alpha \right)^2} = {\left( \alpha \right)^2} \cr & \Rightarrow {\alpha ^2} + 25 - 10\alpha = 0 \cr & \Rightarrow {\left( {\alpha - 5} \right)^2} = 0 \cr & \Rightarrow \alpha = 5 \cr} $$
Then, other root will always real.
Coordinates of the centre of given circle $$ = \left( {\alpha ,\,\alpha } \right)$$
and radius $$ = \sqrt {{{\left( \alpha \right)}^2} + {{\left( \alpha \right)}^2} - {\alpha ^2}} = \sqrt {{\alpha ^2}} = \alpha $$
$$\eqalign{ & \therefore \,{\left( {\alpha - 5} \right)^2} + {\left( \alpha \right)^2} = {\left( \alpha \right)^2} \cr & \Rightarrow {\alpha ^2} + 25 - 10\alpha = 0 \cr & \Rightarrow {\left( {\alpha - 5} \right)^2} = 0 \cr & \Rightarrow \alpha = 5 \cr} $$
Then, other root will always real.