Question

A line parallel to the line $x-3y=2$ touches the circle $x^2+y^2-4x+2y-5=0$ at the point

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

Answer 1

Answer: (B), (C)

$(1,2)$
$(3,-4)$

Circle $s_1:x^2+y^2-4x+2y-5=0$
Line $l:x-3y=2$
Line $l_1$ is parallel to line $l$. Hence, the slope of the required line(s) is $\left(\frac{1}{3}\right)$.
$\therefore$ Line $l_1:y=\frac{1}{3}x+c$
Shift the coordinates system, to $2,-1$ as origin.
In the new coordinate system, equation of
circle, $S:X^2+Y^2=10$
Line $L:X-3Y=1$
Line $L_1:Y=\frac{1}{3}X+C$
Line $L_1$ is tangent to the circle $S$
$\therefore C^2=a^2(1+m^2)$, which gives $C=\pm \frac{10}{3}$
Reverting back to original coordinate system using $X=x-2$ and $Y=y+1$,
equation of line $l_1$ becomes $x-3y=15$ or $x-3y=-5$
Solving $s$ and $l_1$, we get $(x,y)$ as $(1,2)$ or $(3,-4).$
Hence, option B, C.