Question

Consider the circles ${{x}^{2}}+{{y}^{2}}=1\text{ and }{{x}^{2}}+{{y}^{2}}-2x-6y+6=0$ then equation of a common tangent to the two circles is:

& \text{A}.\text{ 4x}-\text{3y}-\text{5}=0 \\\ & \text{B}.\text{ x}+\text{1}=0 \\\ & \text{C}.\text{ 3x}+\text{4y}-\text{5}=0 \\\ & \text{D}.\text{ y}-\text{1}=0 \\\ \end{aligned}$$

Answer 1

At first, use the fact that transverse tangents intersect each other at a point on the line joining the center and will divide the line in the ratio of radii. For direct tangents, their pair point of intersection externally divides the line. Then, use the formula to find slope that is $\dfrac{-pq\pm \sqrt{{{p}^{2}}+{{q}^{2}}-{{R}^{2}}}}{{{R}^{2}}-{{p}^{2}}}$
If the external point is (p, q) to a circle ${{x}^{2}}+{{y}^{2}}={{R}^{2}}$ to get the desired result.

Complete step by step answer:
In the question, we are given equation of two circles ${{x}^{2}}+{{y}^{2}}=1\text{ and }{{x}^{2}}+{{y}^{2}}-2x-6y+6=0$ and we have to find out common tangents in two circles.
At first, we will convert equation of circles and write it in form of ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is center and r is radius.
So we can write ${{x}^{2}}+{{y}^{2}}=1$ as ${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 1 \right)}^{2}}$ so the center of circle is (0, 0) and radius is 1.
Now for equation ${{x}^{2}}+{{y}^{2}}-2x-6y+6=0$ we can write it as,

& {{x}^{2}}-2x+1+{{y}^{2}}-6y+9=0 \\\ & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-3 \right)}^{2}}=4 \\\ & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{2}^{2}} \\\ \end{aligned}$$ So the center is (1, 3) and radius is 2. The distance between centers of the circles is $\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}\Rightarrow \sqrt{{{1}^{2}}+{{3}^{2}}}\Rightarrow \sqrt{10}$ which we get using formula $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ where points are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ The sum of their radius is $1+2\Rightarrow 3$ Thus their distance between centers of circles is greater than the sum of radius. Hence, we will have 4 tangents. Out of 4 tangents two are direct and two are transverse common tangent. The point of intersection of transverse common tangents will intersect each other at points on the line joining the center and will internally divide the line in the ratio of their radius. A point of intersection of direct common tangents will intersect each other at a point on the line joining the center and will externally divide the line in the ratio of their radius. For point of intersection we will use section formula which is: $$\left( \dfrac{cm+an}{m+n},\dfrac{dm+bn}{m+n} \right)$$ If points are (a, b) and (c, d) and are divided in ratio m: n internally. Hence, point of intersection of transverse tangent is $$\left( \dfrac{0\times 2+1\times 1}{3},\dfrac{0\times 2+3\times 1}{3} \right)\Rightarrow \left( \dfrac{1}{3},1 \right)$$ If points are (a, b) and (c d) and are divided in ratio m: n externally, we get points as: $$\left( \dfrac{cm-an}{m-n},\dfrac{dm-bn}{m-n} \right)$$ So the point of intersection of direct tangents are $$\left( \dfrac{0\times 2-1\times 1}{2-1},\dfrac{0\times 2-3\times 1}{2-1} \right)\Rightarrow \left( -1,-3 \right)$$ Now, to find slope of tangent from a external point (p, q) to a circle ${{x}^{2}}+{{y}^{2}}={{R}^{2}}$ we can use a formula, $$\dfrac{-pq\pm \sqrt{{{p}^{2}}+{{q}^{2}}-{{R}^{2}}}}{{{R}^{2}}-{{p}^{2}}}$$ We can find the value of slope from external point (-1,-3) and internal $\left( \dfrac{1}{3},1 \right)$ in both cases R=1. As for (-1,-3) we have ${{R}^{2}}-{{p}^{2}}=0$ thus the value of m is $\infty $ and $\dfrac{-{{q}^{2}}+{{R}^{2}}}{-2Rq}\Rightarrow \dfrac{-9+1}{6}\Rightarrow \dfrac{-4}{3}$ Thus we can find tangents using formula $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is the point and m is the slope. Thus the tangent whose slope is $\infty $ then its slope is $x+1=0$ and whose slope is $\dfrac{-4}{3}$ its line will be $$\begin{aligned} & y+3=\dfrac{-4}{3}\left( x+2 \right) \\\ & \Rightarrow 3y+9=-4x-4 \\\ & \Rightarrow 4x+3y+13=0 \\\ \end{aligned}$$ Similarly for $\left( \dfrac{1}{3},1 \right)$ slopes are given by $$\dfrac{\dfrac{-1}{3}\pm \sqrt{\dfrac{1}{9}+1-1}}{1-\dfrac{1}{9}}\Rightarrow \dfrac{\dfrac{-1}{3}\pm \dfrac{1}{3}}{\dfrac{8}{9}}$$ That is either 0 or $\Rightarrow \dfrac{-3}{4}$ Again using formula to find tangents which is $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ where point is $\left( {{x}_{1}},{{y}_{1}} \right)$ and slope is m. Thus the tangent whose slope is 0 its equation is $y-1=0$ and if slope is $\dfrac{-3}{4}$ then it's equation is $$\begin{aligned} & y-1=\dfrac{-3}{4}\left( x-\dfrac{1}{3} \right) \\\ & \Rightarrow 4y-4=-3x+1 \\\ & \Rightarrow 3x+4y-5=0 \\\ \end{aligned}$$ Hence, all the tangents are $$x+1=0,y-1=0,3x+4y-5=0$$  **So, the correct answers are “Option B, C and D”.** **Note:** Students while finding points using section formulas tend to confuse between internally and externally, so, they should be careful about it. Also, if slope is 0 then it is parallel to x-axis and if slope is $\infty $ then it is parallel to y-axis.