Question

Find the point which is diametrically opposite to the point \[P\left( {1,0} \right)\] on the circle \[{x^2} + {y^2} + 2x + 4y - 3 = 0\].
A. \[\left( {3,4} \right)\]
B. \[\left( {3, - 4} \right)\]
C. \[\left( { - 3,4} \right)\]
D. \[\left( { - 3, - 4} \right)\]

Answer 1

Hint: Simplify the given equation of circle into the standard form. Then using the general equation of the given circle find the center of the circle. In the end, apply the midpoint formula to get the diametrically opposite point of \[P\left( {1,0} \right)\].

Formula Used: The standard form of a circle is: \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\], where \[\left( {h,k} \right)\] is the center and \[r\] is the radius of the circle.
The midpoint of the line joining the points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is: \[\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]

Complete step by step solution:
The given equation of a circle is \[{x^2} + {y^2} + 2x + 4y - 3 = 0\] and the point on the circle is \[P\left( {1,0} \right)\].
Let’s simplify the given equation of the circle into the standard form.
\[{x^2} + {y^2} + 2x + 4y - 3 = 0\]
\[ \Rightarrow \]\[{x^2} + 2x + {y^2} + 4y - 3 = 0\]
\[ \Rightarrow \]\[{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} - 8 = 0\]
\[ \Rightarrow \]\[{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} = {\left( {2\sqrt 2 } \right)^2}\]
Thus, the center of the given circle is \[\left( { - 1, - 2} \right)\].

Image: The circle \[{x^2} + {y^2} + 2x + 4y - 3 = 0\] with diameter passing through the center \[\left( { - 1, - 2} \right)\] and passing through the point \[P\left( {1,0} \right)\]

Let \[Q\left( {x,y} \right)\] be the diametrically opposite point of the point \[P\left( {1,0} \right)\].
Since the center is the mid-point of the endpoints of a diameter.
So, apply midpoint formula to calculate the coordinates of the point \[Q\left( {x,y} \right)\].
\[\left( { - 1, - 2} \right) = \left( {\dfrac{{x + 1}}{2},\dfrac{{y + 0}}{2}} \right)\]
\[ \Rightarrow \]\[\left( { - 1, - 2} \right) = \left( {\dfrac{{x + 1}}{2},\dfrac{y}{2}} \right)\]
Now equate the coordinates. We get
\[ - 1 = \dfrac{{x + 1}}{2}\] and \[ - 2 = \dfrac{y}{2}\]
Solve the above equations.
Let’s solve the equation \[ - 1 = \dfrac{{x + 1}}{2}\].
\[x + 1 = - 1\left( 2 \right)\]
\[ \Rightarrow \]\[x + 1 = - 2\]
\[ \Rightarrow \]\[x = - 2 - 1\]
\[ \Rightarrow \]\[x = - 3\]

Now solve the equation \[ - 2 = \dfrac{y}{2}\].
\[y = - 2\left( 2 \right)\]
\[ \Rightarrow \]\[y = - 4\]
Thus, the diametrically opposite point of the point \[P\left( {1,0} \right)\] is \[Q\left( { - 3, - 4} \right)\].

Hence the correct option is D.

Note: The standard form of a circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\], where \[\left( {h,k} \right)\] is the center and \[r\] is the radius of the circle.
 The general equation of a circle is \[{x^2} + {y^2} + 2gx + 2fy + c = 0\], where \[\left( { - g, - f} \right)\] is the center and \[r = \sqrt {{f^2} + {g^2} - c} \] is the radius of the circle.