Question

Number of common real tangent that can be drawn to the circles $ {x^2} + {y^2} - 2x - 2y = 0 $ and $ {x^2} + {y^2} - 8x - 8y + 14 = 0 $ is
(1) 4
(2) 2
(3) 0
(4) 3

Answer 1

Hint: In order to determine the common tangents between the two circles, we have to find out the distance between their centres and radius. To determine the centres and radius, compare the equation of both circles with the general form of circle i.e. $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $ . The centre will be $ c( - g, - f) $ and radius $ r = \sqrt {{g^2} + {f^2} - C} $ .Find the distance between both the centres using the distance formula $ d = \sqrt {{{\left( {x_2 - x_1} \right)}^2} + {{\left( {y_2 - y_1} \right)}^2}} $ . Clearly you will see that $ {c_1}{c_2} < \left| {{r_1} + {r_2}} \right| $ , so the direct tangent will be 2.

Complete step-by-step answer:
We are given two equations of circles .Let them be $ C_1 $ and $ C_2 $ having equations $ {x^2} + {y^2} - 2x - 2y = 0 $ and $ {x^2} + {y^2} - 8x - 8y + 14 = 0 $ respectively
To find out the number of common tangents between both the circles , we have to first find the distance between the radius and centres of these two circles.
As we know the general form of the equation of a circle is
 $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $
Now comparing each of the equation of circle $ C_1\,and\,C_2 $ with the general form of equation of circle to find the radius and centre
First with $ C1 $
 $ {x^2} + {y^2} - 2x - 2y = 0 $
Since $ 2g $ is the coefficient of $ x $ in the equation so
 $
  2{g_1} = - 2 \\
\Rightarrow {g_1} = - 1 \;
  $
And $ 2f $ is the coefficient of $ y $ in the equation so
 $
  2{f_1} = - 2 \\
\Rightarrow {f_1} = - 1 \;
  $
As we know the centre of a circle is nothing but $ {c_1}( - g, - f) = \left( {1,1} \right) $
And the radius will be $ {r_1} = \sqrt {{g_1}^2 + {f_1}^2 - C} = \sqrt 2 $
Similarly finding the radius and centre of circle $ C_2 $ , we get
 $ {x^2} + {y^2} - 8x - 8y + 14 = 0 $
 $
  2{g_2} = - 8 \\
\Rightarrow {g_2} = - 4 \\
  2{f_2} = - 8 \\
\Rightarrow {f_2} = - 4 \;
  $
SO, the centre $ {c_2}( - {g_2}, - {f_2}) = \left( {4,4} \right) $
And radius $ {r_2} = \sqrt {{g_2}^2 + {f_2}^2 - C} = \sqrt {16 + 16 - 14} = \sqrt {18} = 3\sqrt 2 $
Now finding the distance between $ {c_1} $ and $ {c_2} $ , using distance formula
$ d = \sqrt {{{\left( {x_2 - x_1} \right)}^2} + {{\left( {y_2 - y_1} \right)}^2}} $
$ {c_1}{c_2} = \sqrt {\left( {4 - 1} \right) + \left( {4 - 1} \right)} $
 $
\Rightarrow {r_1} + {r_2} = 3\sqrt 2 + \sqrt 2 \\
\Rightarrow {r_1} + {r_2} = 4\sqrt 2 \;
  $
As we can clearly see, $ {c_1}{c_2} < \left| {{r_1} + {r_2}} \right| $ , so there will be 2 direct common tangents.
Therefore, the answer is option (2).
So, the correct answer is “Option 2”.

Note: 1. When $ \left| {{c_1}{c_2}} \right| > {r_1} + {r_2} $ then there will be two direct and two transverse common tangents.
2. When $ \left| {{c_1}{c_2}} \right| = {r_1} + {r_2} $ , then there will be two direct and one transverse common tangents.
3. If the centre of the circle is not given, then consider it as $ \left( {0,0} \right) $ .