Question
The centre of circle inscribed in square formed by the lines $${x^2} - 8x + 12 = 0$$ and $${y^2} - 14y + 45 = 0,$$ is-
A.
$$\left( {4,\,7} \right)$$
B.
$$\left( {7,\,4} \right)$$
C.
$$\left( {9,\,4} \right)$$
D.
$$\left( {4,\,9} \right)$$
Answer :
$$\left( {4,\,7} \right)$$
Solution :
$$\eqalign{
& {x^2} - 8x + 12 = 0\,\,\, \Rightarrow \left( {x - 6} \right)\left( {x - 2} \right) = 0 \cr
& {y^2} - 14y + 45 = 0\,\,\, \Rightarrow \left( {y - 5} \right)\left( {y - 9} \right) = 0 \cr} $$
Thus sides of square are
$$x = 2,\,\,x = 6,\,\,y = 5,\,\,y = 9$$
Then centre of circle inscribed in square will be
$$\left( {\frac{{2 + 6}}{2},\,\frac{{5 + 9}}{2}} \right) = \left( {4,\,7} \right)$$