Question
The equation of the circle of radius $$2\sqrt 2 $$ whose centre lies on the line $$x - y = 0$$ and which touches the line $$x + y = 4,$$ and whose centre’s coordinates satisfy the inequality $$x + y > 4$$ is :
A.
$${x^2} + {y^2} - 8x - 8y + 24 = 0$$
B.
$${x^2} + {y^2} = 8$$
C.
$${x^2} + {y^2} - 8x + 8y = 24$$
D.
none of these
Answer :
$${x^2} + {y^2} - 8x - 8y + 24 = 0$$
Solution :
$$\eqalign{
& {\text{The centre}} = \left( {t,\,t} \right) \cr
& {\text{The radius}}\,\, = 2\sqrt 2 = \left| {\frac{{t + t - 4}}{{\sqrt 2 }}} \right|.\,\,{\text{So, }}\left| {2t - 4} \right| = 4 \cr
& \therefore \,\,2t - 4 = \pm 4,\,{\text{i}}{\text{.e}}{\text{.,}}\,t = 4,\,0.\,\,{\text{So, the centre}} = \left( {4,\,4} \right),\,\left( {0,\,0} \right) \cr
& \therefore \,\,{\text{satisfying}}\,\,x + y > 4,\,{\text{the centre}} = \left( {4,\,4} \right). \cr} $$