# Question- The number of common tangents to the circle ${x^2} + {y^2} = 4$ and ${x^2} + {y^2} - 8x + 12 = 0$ is

$

{\text{A}}{\text{. 1}} \\

{\text{B}}{\text{. 2}} \\

{\text{C}}{\text{. 3}} \\

{\text{D}}{\text{. 4}} \\

\\

$

Answer

Verified

364.2k+ views

Hint- Here, we will be using the general equation for any circle and the distance formula.

The given equations of two circles are ${x^2} + {y^2} = 4 = {2^2} \to {\text{(1)}}$ and ${x^2} + {y^2} - 8x + 12 = 0{\text{ }} \to {\text{(2)}}$

Since, the general equation of a circle with centre ${\text{C}}\left( {a,b} \right)$ and radius $r$ is given by

${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}{\text{ }} \to {\text{(3)}}$

Equation (2) can be modified in the same form as equation (3) by using completing the square method.

$

{x^2} - 8x + {4^2} - {4^2} + {y^2} + 12 = 0 \Rightarrow \left[ {{x^2} - 8x + {4^2}} \right] - 16 + {y^2} + 12 = 0 \Rightarrow {\left( {x - 4} \right)^2} + {y^2} = 4 = {2^2} \\

\Rightarrow {\left( {x - 4} \right)^2} + {y^2} = {2^2}{\text{ }} \to (4{\text{)}} \\

$

The above equation (4) represents the equation of second circle whose equation is given as equation (2).

On comparing equations (1) and (3), we get

${a_1} = 0,{\text{ }}{b_1} = 0$ and ${r_1} = 2$ where centre of the first circle whose equation is given by equation (1) is ${{\text{C}}_1}\left( {{a_1} = 0,{b_1} = 0} \right) \Leftrightarrow {{\text{C}}_1}\left( {0,0} \right)$ and radius ${r_1} = 2$.

On comparing equations (3) and (4), we get

${a_2} = 4,{\text{ }}{b_2} = 0$ and ${r_2} = 2$ where centre of the second circle whose equation is given by equation (2) is ${{\text{C}}_2}\left( {{a_2} = 4,{b_2} = 0} \right) \Leftrightarrow {{\text{C}}_2}\left( {4,0} \right)$ and radius is ${r_2} = 2$.

Now, the distance between the centres of two circles is evaluated using distance formula

${\text{d = }}{{\text{C}}_1}{{\text{C}}_2} = \sqrt {{{\left( {{a_2} - {a_1}} \right)}^2} + {{\left( {{b_2} - {b_1}} \right)}^2}} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = 4$.

Here, ${r_1} + {r_2} = 2 + 2 = 4$

Clearly, ${{\text{C}}_1}{{\text{C}}_2} = {r_1} + {r_2}$ which means that the given two circles are touching each other externally.

Also, in total there are three common tangents which can be drawn to the given two circles which are touching each other externally.

Therefore, option C is correct.

Note- These type of problems are solved by considering the general equation of the circle and then comparing the given equations to find the given circle’s dimensions and then evaluating the centre to centre distance between the given circles.

The given equations of two circles are ${x^2} + {y^2} = 4 = {2^2} \to {\text{(1)}}$ and ${x^2} + {y^2} - 8x + 12 = 0{\text{ }} \to {\text{(2)}}$

Since, the general equation of a circle with centre ${\text{C}}\left( {a,b} \right)$ and radius $r$ is given by

${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}{\text{ }} \to {\text{(3)}}$

Equation (2) can be modified in the same form as equation (3) by using completing the square method.

$

{x^2} - 8x + {4^2} - {4^2} + {y^2} + 12 = 0 \Rightarrow \left[ {{x^2} - 8x + {4^2}} \right] - 16 + {y^2} + 12 = 0 \Rightarrow {\left( {x - 4} \right)^2} + {y^2} = 4 = {2^2} \\

\Rightarrow {\left( {x - 4} \right)^2} + {y^2} = {2^2}{\text{ }} \to (4{\text{)}} \\

$

The above equation (4) represents the equation of second circle whose equation is given as equation (2).

On comparing equations (1) and (3), we get

${a_1} = 0,{\text{ }}{b_1} = 0$ and ${r_1} = 2$ where centre of the first circle whose equation is given by equation (1) is ${{\text{C}}_1}\left( {{a_1} = 0,{b_1} = 0} \right) \Leftrightarrow {{\text{C}}_1}\left( {0,0} \right)$ and radius ${r_1} = 2$.

On comparing equations (3) and (4), we get

${a_2} = 4,{\text{ }}{b_2} = 0$ and ${r_2} = 2$ where centre of the second circle whose equation is given by equation (2) is ${{\text{C}}_2}\left( {{a_2} = 4,{b_2} = 0} \right) \Leftrightarrow {{\text{C}}_2}\left( {4,0} \right)$ and radius is ${r_2} = 2$.

Now, the distance between the centres of two circles is evaluated using distance formula

${\text{d = }}{{\text{C}}_1}{{\text{C}}_2} = \sqrt {{{\left( {{a_2} - {a_1}} \right)}^2} + {{\left( {{b_2} - {b_1}} \right)}^2}} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = 4$.

Here, ${r_1} + {r_2} = 2 + 2 = 4$

Clearly, ${{\text{C}}_1}{{\text{C}}_2} = {r_1} + {r_2}$ which means that the given two circles are touching each other externally.

Also, in total there are three common tangents which can be drawn to the given two circles which are touching each other externally.

Therefore, option C is correct.

Note- These type of problems are solved by considering the general equation of the circle and then comparing the given equations to find the given circle’s dimensions and then evaluating the centre to centre distance between the given circles.

Last updated date: 27th Sep 2023

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