Question

The equation of the circle described on the chord \[3x + y + 5 = 0\] of the circle \[{x^2} + {y^2} = 16\] as diameter is:
A) \[{x^2} + {y^2} + 3x + y - 11 = 0\]
B) \[{x^2} + {y^2} + 3x + y + 1 = 0\]
C) \[{x^2} + {y^2} + 3x + y - 2 = 0\]
D) \[{x^2} + {y^2} + 3x + y - 22 = 0\]

Answer 1

Hint: Here first we will find the end point of the chord by solving the equations of the chord and the given circle and then use it as a diameter of the another circle and find the midpoint of the chord which will be the center of the another circle and then find the distance between the center and one of the end points of the chord to get the radius of the another circle and then finally find its equation.
The equation of a circle with center (a, b) and radius r is given by:-
\[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\]

Complete step-by-step answer:
The given chord is:-
\[3x + y + 5 = 0\]…………………….(1)
The given equation of the circle is:-
\[{x^2} + {y^2} = 16\]………………(2)
Now we will find the endpoints of the chord.
In order to find the endpoints we need to find the intersection points of chord and the circle
Hence solving both the equations we get:-
From equation 1 we get:-
\[y = - 3x - 5\]
Putting this value in equation 2 we get:-
\[{x^2} + {\left( { - 3x - 5} \right)^2} = 16\]
Simplifying it we get:-
\[
  {x^2} + {\left( - \right)^2}{\left( {3x + 5} \right)^2} = 16 \\
   \Rightarrow {x^2} + {\left( {3x + 5} \right)^2} = 16 \\
 \]
Now we know that:
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Hence applying this formula we get:-
\[{x^2} + {\left( {3x} \right)^2} + {\left( 5 \right)^2} + 2\left( {3x} \right)\left( 5 \right) = 16\]
Solving it further we get:-
\[
  {x^2} + 9{x^2} + 25 + 30x = 16 \\
   \Rightarrow 10{x^2} + 30x + 25 - 16 = 0 \\
 \]
Simplifying it further we get:-
\[10{x^2} + 30x + 9 = 0\]
Now we will do this equation using the quadratic formula.
The quadratic formula for standard equation of form \[a{x^2} + bx + c = 0\] is given by:-
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Hence applying this formula for above equation we get:-
\[x = \dfrac{{ - \left( {30} \right) \pm \sqrt {{{\left( {30} \right)}^2} - 4\left( {10} \right)\left( 9 \right)} }}{{2\left( {10} \right)}}\]
Solving it further we get:-
\[
  x = \dfrac{{ - 30 \pm \sqrt {900 - 360} }}{{20}} \\
   \Rightarrow x = \dfrac{{ - 30 \pm \sqrt {540} }}{{20}} \\
 \]
Taking square root we get:-
\[
  x = \dfrac{{ - 30 \pm 6\sqrt {15} }}{{20}} \\
   \Rightarrow x = \dfrac{{ - 15 \pm 3\sqrt {15} }}{{10}} \\
 \]
Now putting this value in equation 1 we get:-
\[3\left( {\dfrac{{ - 15 \pm 3\sqrt {15} }}{{10}}} \right) + y + 5 = 0\]
Solving for y we get:-
\[y = - 5 - 3\left( {\dfrac{{ - 15 \pm 3\sqrt {15} }}{{10}}} \right)\]
Solving it further we get:-
\[
  y = - 5 - 3\left[ {\dfrac{{ - 15 + 3\sqrt {15} }}{{10}}} \right], - 5 - 3\left[ {\dfrac{{ - 15 - 3\sqrt {15} }}{{10}}} \right] \\
   \Rightarrow y = \dfrac{{ - 5 - 9\sqrt {15} }}{{10}},\dfrac{{ - 5 + 9\sqrt {15} }}{{10}} \\
 \]
Hence the end points of the chord AB are:-
\[\left[ {\dfrac{{ - 15 + 3\sqrt {15} }}{{10}},\dfrac{{ - 5 - 9\sqrt {15} }}{{10}}} \right],\left[ {\dfrac{{ - 15 - 3\sqrt {15} }}{{10}},\dfrac{{ - 5 + 9\sqrt {15} }}{{10}}} \right]\]
Now it is given that chord AB is diameter for another circle
Hence center of the circle would be the midpoint of the chord AB
The midpoint of two points \[\left( {{x_1},{y_1}} \right)\& \left( {{x_2},{y_2}} \right)\]is given by:-
\[x = \dfrac{{{x_1} + {x_2}}}{2},y = \dfrac{{{y_1} + {y_2}}}{2}\]
Hence the midpoint of chord AB is:-
\[x = \dfrac{{\dfrac{{ - 15 + 3\sqrt {15} }}{{10}} + \dfrac{{ - 15 - 3\sqrt {15} }}{{10}}}}{2},y = \dfrac{{\dfrac{{ - 5 - 9\sqrt {15} }}{{10}} + \dfrac{{ - 5 + 9\sqrt {15} }}{{10}}}}{2}\]
Taking LCM we get:-
\[x = \dfrac{{\dfrac{{ - 15 + 3\sqrt {15} - 15 - 3\sqrt {15} }}{{10}}}}{2},y = \dfrac{{\dfrac{{ - 5 - 9\sqrt {15} - 5 + 9\sqrt {15} }}{{10}}}}{2}\]
Now solving them further we get:-
\[
  x = \dfrac{{ - 30}}{{20}},y = \dfrac{{ - 10}}{{20}} \\
   \Rightarrow x = \dfrac{{ - 3}}{2},y = \dfrac{{ - 1}}{2} \\
 \]
Hence the center of the circle is:-
\[\left( {a,b} \right) = \left( { - \dfrac{3}{2}, - \dfrac{1}{2}} \right)\]………………………………(3)
Now we need to find the radius of the circle.
Now radius will be the distance between the center and the endpoint \[\left[ {\dfrac{{ - 15 + 3\sqrt {15} }}{{10}},\dfrac{{ - 5 - 9\sqrt {15} }}{{10}}} \right]\]
The distance between two points \[\left( {{x_1},{y_1}} \right)\& \left( {{x_2},{y_2}} \right)\] is given by:-
\[d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \]
Therefore the distance between the center and the endpoint \[\left[ {\dfrac{{ - 15 + 3\sqrt {15} }}{{10}},\dfrac{{ - 5 - 9\sqrt {15} }}{{10}}} \right]\]is:-
\[r = \sqrt {{{\left( {\dfrac{{ - 3}}{2} - \dfrac{{ - 15 + 3\sqrt {15} }}{{10}}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{2} - \dfrac{{ - 5 - 9\sqrt {15} }}{{10}}} \right)}^2}} \]
Squaring both the sides we get:-
\[
  {r^2} = {\left[ {\sqrt {{{\left( {\dfrac{{ - 3}}{2} - \dfrac{{ - 15 + 3\sqrt {15} }}{{10}}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{2} - \dfrac{{ - 5 - 9\sqrt {15} }}{{10}}} \right)}^2}} } \right]^2} \\
   \Rightarrow {r^2} = {\left( {\dfrac{{ - 3}}{2} - \dfrac{{ - 15 + 3\sqrt {15} }}{{10}}} \right)^2} + {\left( {\dfrac{{ - 1}}{2} - \dfrac{{ - 5 - 9\sqrt {15} }}{{10}}} \right)^2} \\
 \]
Taking LCM we get:-
\[
  {r^2} = {\left( {\dfrac{{ - 3\left( 5 \right) - \left[ { - 15 + 3\sqrt {15} } \right]}}{{10}}} \right)^2} + {\left( {\dfrac{{ - 1\left( 5 \right) - \left[ { - 5 - 9\sqrt {15} } \right]}}{{10}}} \right)^2} \\
    \\
 \]
Solving it further we get:-
\[
  {r^2} = {\left( {\dfrac{{ - 15 + 15 - 3\sqrt {15} }}{{10}}} \right)^2} + {\left( {\dfrac{{ - 5 + 5 + 9\sqrt {15} }}{{10}}} \right)^2} \\
   \Rightarrow {r^2} = {\left( {\dfrac{{ - 3\sqrt {15} }}{{10}}} \right)^2} + {\left( {\dfrac{{9\sqrt {15} }}{{10}}} \right)^2} \\
 \]
Evaluating the squares we get:-
\[{r^2} = \dfrac{{9 \times 15}}{{100}} + \dfrac{{81 \times 15}}{{100}}\]
Taking \[\dfrac{{15}}{{100}}\] as common we get:-
\[{r^2} = \dfrac{{15}}{{100}}\left[ {9 + 81} \right]\]
Calculating it further we get:-
\[
  {r^2} = \dfrac{{15}}{{100}}\left[ {90} \right] \\
   \Rightarrow {r^2} = \dfrac{{27}}{2}...............................\left( 4 \right) \\
 \]
Now we know that
Now putting the equation of a circle with center (a, b) and radius r is given by:-
\[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\]
Hence putting the values from equation3 and 4 we get:-
\[{\left( {x - \left( {\dfrac{{ - 3}}{2}} \right)} \right)^2} + {\left( {y - \left( {\dfrac{{ - 1}}{2}} \right)} \right)^2} = \dfrac{{27}}{2}\]
Applying the following identity:-
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Solving it further we get:-
\[
  {x^2} + {\left( {\dfrac{{ - 3}}{2}} \right)^2} - 2x\left( {\dfrac{{ - 3}}{2}} \right) + {y^2} + {\left( {\dfrac{{ - 1}}{2}} \right)^2} - 2y\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{{27}}{2} \\
  {x^2} + \dfrac{9}{4} + 6x + {y^2} + \dfrac{1}{4} + y = \dfrac{{27}}{2} \\
 \]
Taking LCM we get:-
\[
  \dfrac{{4{x^2} + 4{y^2} + 12x + 4y + 9 + 1}}{4} = \dfrac{{27}}{2} \\
   \Rightarrow \dfrac{{4{x^2} + 4{y^2} + 12x + 4y + 9 + 1}}{2} = 27 \\
 \]
Simplifying it further we get:-
\[
  4{x^2} + 4{y^2} + 12x + 4y + 10 = 27 \times 2 \\
   \Rightarrow 4{x^2} + 4{y^2} + 12x + 4y + 10 = 54 \\
   \Rightarrow 4{x^2} + 4{y^2} + 12x + 4y + 10 - 54 = 0 \\
   \Rightarrow 4{x^2} + 4{y^2} + 12x + 4y - 44 = 0 \\
 \]
Dividing the equation by 4 we get:-
\[{x^2} + {y^2} + 3x + y - 11 = 0\]
Hence the equation of circle is:
\[{x^2} + {y^2} + 3x + y - 11 = 0\]

Hence option A is the correct option.

Note: Students should take note that the center of a circle is the midpoint of the diameter and radius is half of the length of diameter.
Also, chord is the line segment joining two points on the circumference of the circle and diameter is the special case of chord of a circle which passes through its center.
Students might make mistakes in calculations.
Students can also use the following general form of equation of circle:
\[{x^2} + {y^2} + 2gx + 2fy + c = 0\] where \[center \equiv \left( { - g, - f} \right)\] and radius \[r = \sqrt {{g^2} + {f^2} - c} \]