Question

The common tangent to the circles ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 4}}$and ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + 6x + 8y - 24 = 0}}$also passes through the point:
A. $\left( {{\text{ - 4,6}}} \right)$
B. $\left( {{\text{6, - 2}}} \right)$
C. $\left( {{\text{ - 6,4}}} \right)$
D. $\left( {{\text{4, - 2}}} \right)$

Answer 1

Hint: We have equations of 2 circles. We can find the center and radius of both circles from the equation. Then we can find whether the circles meet internally or externally. Then we can use suitable equations to find the equation of the tangent and check whether the given points satisfy the tangent of the equation.

Complete step by step answer:

We have equations of two circles. ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 4}}$ and ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + 6x + 8y - 24 = 0}}$.
From the equation of a circle in its standard form \[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\], we can identify the r is the radius and \[(a,b)\] is the centre of the circle.
Let’s consider the 1st circle ${x^2} + {y^2} = 4$. As the equation is in the standard form, we get center and radius as,
$
  {{\text{C}}_{\text{1}}}{\text{ = (0,0)}} \\
  {{\text{r}}_{\text{1}}}{\text{ = 2}} \\
 $
For the 2nd circle, \[{x^2} + {y^2} + 6x + 8y - 24 = 0\]
Simplifying the equation, we get,
\[
  {x^2} + {y^2} + \left( {2 \times 3 \times x} \right) + \left( {2 \times 4 \times y} \right) + {3^2} + {4^2} - 9 - 16 - 24 = 0 \\
  {\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} = 49 \\
 \]
Now we have the equation in standard form. So, we get center and radius as,
$
  {{\text{C}}_{\text{2}}}{\text{ = ( - 3, - 4)}} \\
  {{\text{r}}_{\text{2}}}{\text{ = 7}} \\
 $
Now we can check whether the circle meets internally. For 2 circles to meet internally, the distance between their centers must be equal to the difference in their radius.
Distance between the centers is given by,
We use distance formula which is \[\sqrt {{{{\text{(}}{{\text{x}}_{\text{2}}}{\text{ - }}{{\text{x}}_{\text{1}}}{\text{)}}}^{\text{2}}}{\text{ + (}}{{\text{y}}_{\text{2}}}{\text{ - }}{{\text{y}}_{\text{1}}}{{\text{)}}^{\text{2}}}} \]
\[
  {C_1}{C_2} = \sqrt {{{\left( {0 + 3} \right)}^2} + {{\left( {0 + 4} \right)}^2}} \\
   = \sqrt {9 + 16} \\
   = \sqrt {25} \\
   = 5units \\
 \]
Difference in radius is given by,
${{\text{r}}_{\text{2}}}{\text{ - }}{{\text{r}}_{\text{1}}}{\text{ = 7 - 2 = 5units}}$
As the difference in radius and distance between the center is equal, the circle meets internally.
Then the equation of the common tangent is given by the difference of the equations of the circle.
${S_2} - {S_1} = 0$
Substituting the equations of the circles to the equation of common tangent, we get,
\[
  \left( {{x^2} + {y^2} + 6x + 8y - 24} \right) - \left( {{x^2} + {y^2} - 4} \right) = 0 \\
   \Rightarrow 6x + 8y - 20 = 0 \\
   \Rightarrow 3x + 4y - 10 = 0 \\
 \]
Now we can check whether the given points satisfy in the equation of common tangent.
Option 1 is $\left( {{\text{ - 4,6}}} \right)$ $ \Rightarrow 3x + 4y - 10 = 3\left( { - 4} \right) + 4\left( 6 \right) - 10 = - 12 + 24 - 10 = 2 \ne 0$
So, $\left( {{\text{ - 4,6}}} \right)$is not a point on the common tangent.
Option 2 is $\left( {{\text{6, - 2}}} \right)$$ \Rightarrow 3x + 4y - 10 = 3\left( 6 \right) + 4\left( { - 2} \right) - 10 = 18 - 8 - 10 = 0$
So, $\left( {{\text{6, - 2}}} \right)$ is a point on the common tangent
Option 3 is$\left( {{\text{ - 6,4}}} \right)$$ \Rightarrow {\text{3x + 4y - 10 = 3}}\left( {{\text{ - 6}}} \right){\text{ + 4}}\left( {\text{4}} \right){\text{ - 10 = - 18 + 8 - 10 = - 20}} \ne {\text{0}}$.
And option 4 is$\left( {{\text{4, - 2}}} \right)$$ \Rightarrow {\text{3x + 4y - 10 = 3}}\left( {\text{4}} \right){\text{ + 4}}\left( {{\text{ - 2}}} \right){\text{ - 10 = 12 - 8 - 10 = - 6}} \ne {\text{0}}$
So, the points $\left( {{\text{ - 6,4}}} \right)$and $\left( {{\text{4, - 2}}} \right)$are not points on the tangent.
Therefore, the correct answer is option B.

Note: The center and radius of a circle can be found out from the equation of the circle in its standard form i.e. \[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\], where the Centre is \[(a,b)\], and the radius is r. The two circles meet internally if the distance between their centers is equal to the difference in their radii. Then the equation for its common tangent is the difference of the equation of the 2 circles. If a point lies on a line, the point will satisfy the equation of the line. We cannot find the solution without the options. We cannot solve the equation of the common tangent to find the points as we have only one equation. While substituting the points, we must take care of the sign of the coordinates. Make sure that the equation is in the standard form to check whether the points satisfy the equation.