Question

The point on the straight line $x+y=2$ which is nearest to the circle $x^2+y^2-10x+2y+22=0$

• A. $(4,0)$
• B. $(4-2)$
• C. $(-2,9)$
• D. $(3,-1)$

Answer 1

The correct option is A $(4,0)$

The point nearer to the circle is a foot of the perpendicular from the center

which is $\frac{(h-x)}{a}=\frac{(k-y)}{b}=\frac{-(ax+by+c)}{a^2+b^2}\cdots \left(1\right)$

$(x,y)=(5,-1)$

$a=1,b=1,c=-2$

Put these values in $\left(1\right)$

$⟹h=4,k=0$

so $(4,0)$ is nearest point.