Question

I : The equations to the direct common tangents to the circles $x^2+y^2+6x+4y+4=0$, $x^2+y^2-2x=0$ are $y-1=0$, $4x-3y-9=0$
II : The equations to the transverse common tangents to the circles $x^2+y^2-4x-10y+28=0$, $x^2+y^2+4x-6y+4=0$ are $x-1=0,3x+4y-21=0$

• A. Only I is true
• B. Only II is true
• C. both I & II are true
• D. neither I nor II true

Answer 1

Answer: (C)

both I & II are true

$S_1:x^2+y^2+6x+4y+4=0,C_1=(-3,-2),r_1=3$

$S_2:x^2+y^2–2x=0,C_2=(1,0),r_2=1$

The external center of similitude cuts the line joining the centers in the ratio $r_1:r_2$ externally

$P=\frac{r_1{C}_1–r_2{C}_2}{r_1–r_2}$

$=(\frac{3–(-3)}{2},\frac{0-(-2)}{2})=(3,1)$

Any tangent to $S_2$ is given by

$y=m(x–1)+\sqrt{m^2+1}–\left(1\right)$

$(3,1)$ passes through the above line

$⟹1=m(3–1)+\sqrt{m^2+1}$

$4m^2+1–4m=m^2+1$

$3m^2=4m$

$m=0,\frac{4}{3}$

Substituting in (1)

The common tangents are

$y=1,4x–3y-9$

(II)

$S_1:(x-2{)}^2+(y–5{)}^2=1,C_1=(2,5),r_1=1$

$S_2:(x+2{)}^2+(y–3{)}^2=9,C_2=(-2,3),r_2=3$

Any tangent to $S_1$ is given by

$(y–5)=m(x–2)\pm 1\sqrt{m^2+1}$

$y=mx+(5\pm \sqrt{m^2+1}–2m)–\left(1\right)$

Any tangent to $S_2$ is given by

$(y–3)=m(x+2)\pm 3\sqrt{m^2+1}$

$y=mx+(3+2m\pm 3\sqrt{m^2+1})–\left(2\right)$

(1) And (2) represent the same line

$⟹5\pm \sqrt{m^2+1}–2m=3+2m\pm 3\sqrt{m^2+1}$

$4m–2=-2\sqrt{m^2+1}$

$16m^2+4–8m=2m^2+2$

$8m^2–8m+2=0$

$⟹m=\frac{1}{2}$

(or)

$4m–2=4\sqrt{m^2+1}$

$⟹m=\frac{-3}{4}$

The internal central of similitude is given by

$Q=\frac{r_1{C}_2+r_2{C}_1}{r_1+r_2}=(\frac{4}{4},\frac{18}{4})=(1,\frac{9}{2})$

Using (1) and verifying slopes substituting Q we get the common tangents as

$x=1,3x+4y–21=0$