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Question: The common tangent to the circles \({x^2} + {y^2} = 4\) and \({x^2} + {y^2} + 6x + 8y - 24 = 0\) als...

The common tangent to the circles x2+y2=4{x^2} + {y^2} = 4 and x2+y2+6x+8y−24=0{x^2} + {y^2} + 6x + 8y - 24 = 0 also passes through the point:
A. (−4,6)\left( { - 4,6} \right)
B. (6,−2)\left( {6, - 2} \right)
C. (−6,4)\left( { - 6,4} \right)
D. (4,−2)\left( {4, - 2} \right)

Explanation

Solution

We are given two equations of circles. We can find the radius and the centre of the circles by comparing the equation with the standard equation of circles. Then, we can use equations of tangent and the given points if they satisfy the equation of the tangent.

Complete step by step solution:
We have the equations of circle as x2+y2=4{x^2} + {y^2} = 4 and x2+y2+6x+8y−24=0{x^2} + {y^2} + 6x + 8y - 24 = 0
The standard equation of the circle is (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}, where (h,k)\left( {h,k} \right) are the coordinates of centre of a circle and rr is the radius of the circle.
Hence, for the equation of the circle, x2+y2=4{x^2} + {y^2} = 4, we have centre as (0,0)\left( {0,0} \right) and radius is 2 units.
The equation of circle, x2+y2+6x+8y−24=0{x^2} + {y^2} + 6x + 8y - 24 = 0 can be written as
x2+y2+2×3×x+2×4×y=24{x^2} + {y^2} + 2 \times 3 \times x + 2 \times 4 \times y = 24
We will complete the squares and then compare it with the standard equation.
x2+y2+2×3×x+2×4×y+9+16=24+9+16 ⇒(x+3)2+(y+4)2=49  {x^2} + {y^2} + 2 \times 3 \times x + 2 \times 4 \times y + 9 + 16 = 24 + 9 + 16 \\\ \Rightarrow {\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} = 49 \\\
Hence, the coordinates of the centre of the circle is (−3,−4)\left( { - 3, - 4} \right) and the radius is 7 units.
Now, let us check if the circles meet internally or not.
If these circles meet internally, then the distance between their centres must be equal to or less than the difference between their radii.
We will use a distance formula to find the distance between the centres.
The distance formula states that if (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) are two points, then the distance between these two points is (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
(0+3)2+(0+4)2 =9+16 =25 =5  \sqrt {{{\left( {0 + 3} \right)}^2} + {{\left( {0 + 4} \right)}^2}} \\\ = \sqrt {9 + 16} \\\ = \sqrt {25} \\\ = 5 \\\
The difference between the radii of two circles is 7−2=57 - 2 = 5
This implies that the circles meet internally and the equation of common tangent is given by difference of the equations of the circles.
(x2+y2+6x+8y−24)−(x2+y2−4)=0 ⇒6x+8y−20=0 ⇒3x+4y−10=0  \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 \\\
Substitute the points from option A , that is (−4,6)\left( { - 4,6} \right)
3(−4)+4(6)−10=2≠03\left( { - 4} \right) + 4\left( 6 \right) - 10 = 2 \ne 0
Substitute the points from option B , that is (6,−2)\left( {6, - 2} \right)
3(6)+4(−2)−10=03\left( 6 \right) + 4\left( { - 2} \right) - 10 = 0
Substitute the points from option C , that is (−6,4)\left( { - 6,4} \right)
3(−6)+4(4)−10=−20≠03\left( { - 6} \right) + 4\left( 4 \right) - 10 = - 20 \ne 0
Substitute the points from option D, that is (4,−2)\left( {4, - 2} \right)
3(4)+4(−2)−10=−6≠03\left( 4 \right) + 4\left( { - 2} \right) - 10 = - 6 \ne 0

Hence, option B is correct.

Note:
The standard equation of the circle is (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}, where (h,k)\left( {h,k} \right) are the coordinates of centre of a circle and rr is the radius of the circle. This question cannot be done without given points in the option. The point which satisfies the formed equation of the common tangent is the required point.