• # question_answer Suppose the straight line $x+y=5$touches the circle ${{x}^{2}}+{{y}^{2}}-2x-4y+3=0$. Then, the coordinates of the point of contact are A)  $(3,2)$           B)  $(2,3)$ C)  $(4,1)$ D)  $(1,4)$

Given equation of circle, ${{x}^{2}}+{{y}^{2}}-2x-4y+3=0$ ?.(i) and equation of straight line, $x+y=5$ ?.(ii) On solving Eqs. (i) and (ii), we get ${{x}^{2}}+{{(5-x)}^{2}}-2x-4(5-x)+3=0$ $\Rightarrow$ $2{{x}^{2}}-8x+8=0$ $\Rightarrow$ ${{x}^{2}}-4x+4=0$ $\Rightarrow$ ${{(x-2)}^{2}}=0$ $\Rightarrow$ $x=2$ From Eq. (ii), $y=3$ So, the point of contact is $(2,3)$.