Question

Find the equations of the straight lines which touch both the circles ${x^2} + {y^2} = 4$ and ${\left( {x - 4} \right)^2} + {y^2} = 1$.

Answer 1

First of all, find the centre and radius of the given circles and find the internal and external centre of similitudes. Then find the pair of transverse common tangents and direct common tangents to the given circles. Use this concept to reach the solution.

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Complete step-by-step answer:

Given circle equations are ${x^2} + {y^2} = 4$ and ${\left( {x - 4} \right)^2} + {y^2} = 1$

We know that for the circle equation ${x^2} + {y^2} = {a^2}$ the centre of the circle is $\left( {0,0} \right)$ and radius of the circle is $a$.

So, for the circle equation ${x^2} + {y^2} = 4$ the centre is ${C_1}\left( {0,0} \right)$ and radius is ${r_1} = 2$.

We know that for the circle equation ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}$ the centre of the circle is $\left( {h,k} \right)$ and radius of the circle is $a$.

So, for the circle equation ${\left( {x - 4} \right)^2} + {y^2} = 1$ the centre is ${C_2}\left( {4,0} \right)$ and radius is ${r_2} = 1$.

We know that the distance between the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.

Now, consider the distance between the centres of the two circles i.e.,

${C_1}{C_2} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = 4$

As ${C_1}{C_2} > {r_1} + {r_2} \Rightarrow 4 > 2 + 1$, there will be four common tangents to the circles as shown in the below figure:

We know that if a point $C$ divides the two points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ in the ratio $p:q$ the point $C$ is given by $C = \left( {\dfrac{{p{x_2} + q{x_1}}}{{p + q}},\dfrac{{p{y_2} + q{y_1}}}{{p + q}}} \right)$.

The internal centre of similitude, ${A_1}$ divides ${C_1}{C_2}$ in the ratio $2:1$ internally.

$$\Rightarrow {A_1} = \left( {\dfrac{{2\left( 4 \right) + 1\left( 0 \right)}}{{2 + 1}},\dfrac{{2\left( 0 \right) + 1\left( 0 \right)}}{{2 + 1}}} \right) \\\ \therefore {A_1} = \left( {\dfrac{8}{3},0} \right) \\\$$

We know that the equation to the pair of transverse common tangents at point $\left( {h,k} \right)$ to the circle ${x^2} + {y^2} = {a^2}$ is given by ${\left( {xh + yk - {a^2}} \right)^2} = \left( {{h^2} + {k^2} - {a^2}} \right)\left( {{x^2} + {y^2} - {a^2}} \right)$

So, equation to the pair of transverse common tangents at $\left( {\dfrac{8}{3},0} \right)$ to the circle ${x^2} + {y^2} = 4$ is

$$\Rightarrow {\left( {x\left( {\dfrac{8}{3}} \right) + y\left( 0 \right) - 4} \right)^2} = \left( {{{\left( {\dfrac{8}{3}} \right)}^2} + {{\left( 0 \right)}^2} - 4} \right)\left( {{x^2} + {y^2} - 4} \right) \\\ \Rightarrow {\left( {\dfrac{{8x}}{3} - 4} \right)^2} = \left( {\dfrac{{64}}{9} - 4} \right)\left( {{x^2} + {y^2} - 4} \right) \\\$$

Opening the terms inside the brackets, we have

$$\Rightarrow \dfrac{{64{x^2}}}{9} - 2\left( 4 \right)\left( {\dfrac{{8x}}{3}} \right) + 16 = \left( {\dfrac{{64 - 36}}{9}} \right)\left( {{x^2} + {y^2} - 4} \right) \\\ \Rightarrow \dfrac{{64{x^2}}}{9} - \dfrac{{64x}}{3} + 16 = \left( {\dfrac{{28}}{9}} \right)\left( {{x^2} + {y^2} - 4} \right) \\\ \Rightarrow \dfrac{{64{x^2} - 192x + 144}}{9} = \dfrac{{28{x^2} + 28{y^2} - 112}}{9} \\\$$

Cancelling and grouping the terms, we have

$$\Rightarrow 64{x^2} - 28{x^2} - 28{y^2} - 192x + 144 + 112 = 0 \\\ \Rightarrow 36{x^2} - 192x + 256 = 28{y^2} \\\ \Rightarrow {\left( {6x - 16} \right)^2} = 28{y^2} \\\$$

Rooting on both sides, we get

$$\Rightarrow 6x - 16 = \pm \sqrt {28} y \\\ \Rightarrow 6x - 16 = \pm 2\sqrt 7 y \\\ \therefore 3x \pm \sqrt 7 y - 8 = 0 \\\$$

The external centre of similitude, ${A_2}$ divides ${C_1}{C_2}$ in the ratio $2:1$ externally.

$$\Rightarrow {A_2} = \left( {\dfrac{{2\left( 4 \right) - 1\left( 0 \right)}}{{2 - 1}},\dfrac{{2\left( 0 \right) - 1\left( 0 \right)}}{{2 - 1}}} \right) \\\ \therefore {A_2} = \left( {8,0} \right) \\\$$

We know that the equation to the pair of direct common tangents at point $\left( {h,k} \right)$ to the circle ${x^2} + {y^2} = {a^2}$ is given by ${\left( {xh + yk - {a^2}} \right)^2} = \left( {{h^2} + {k^2} - {a^2}} \right)\left( {{x^2} + {y^2} - {a^2}} \right)$

So, equation to the pair of direct common tangents at $\left( {8,0} \right)$ to the circle ${x^2} + {y^2} = 4$ is

$$\Rightarrow {\left( {x\left( 8 \right) + y\left( 0 \right) - 4} \right)^2} = \left( {{{\left( 8 \right)}^2} + {{\left( 0 \right)}^2} - 4} \right)\left( {{x^2} + {y^2} - 4} \right) \\\ \Rightarrow {\left( {8x - 4} \right)^2} = \left( {64 - 4} \right)\left( {{x^2} + {y^2} - 4} \right) \\\$$

Opening the terms inside the brackets, we have

$$\Rightarrow 64{x^2} - 2\left( 4 \right)\left( {8x} \right) + 16 = \left( {60} \right)\left( {{x^2} + {y^2} - 4} \right) \\\ \Rightarrow 64{x^2} - 64x + 16 = 60{x^2} + 60{y^2} - 240 \\\ \Rightarrow 64{x^2} - 60{x^2} - 60{y^2} - 64x + 16 + 240 = 0 \\\ \Rightarrow 4{x^2} - 60{y^2} - 64x + 256 = 0 \\\$$

Grouping and cancelling the common terms, we have

$$\Rightarrow 4\left( {{x^2} - 15{y^2} - 16x + 64} \right) = 0 \\\ \Rightarrow {x^2} - 16x + 64 = 15{y^2} \\\ \Rightarrow {\left( {x - 8} \right)^2} = 15{y^2} \\\$$

Rooting on both sides, we get

$$\Rightarrow x - 8 = \pm \sqrt {15} y \\\ \therefore x \pm \sqrt {15} y - 8 = 0 \\\$$

Thus, the equation of straight lines is $3x + \sqrt 7 y - 8 = 0,3x - \sqrt 7 y - 8 = 0,x + \sqrt {15} y - 8 = 0{\text{ and }}x - \sqrt {15} y - 8 = 0$.

Note: The equation to the pair of transverse common tangents at point $\left( {h,k} \right)$ to the circle ${x^2} + {y^2} = {a^2}$ is given by ${\left( {xh + yk - {a^2}} \right)^2} = \left( {{h^2} + {k^2} - {a^2}} \right)\left( {{x^2} + {y^2} - {a^2}} \right)$. The equation to the pair of direct common tangents at point $\left( {h,k} \right)$ to the circle ${x^2} + {y^2} = {a^2}$ is given by ${\left( {xh + yk - {a^2}} \right)^2} = \left( {{h^2} + {k^2} - {a^2}} \right)\left( {{x^2} + {y^2} - {a^2}} \right)$.