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A circle with centre C has equation $x^2 + y^2 + 8x - 12y = 12$ - AQA - A-Level Maths: Mechanics - Question 11 - 2017 - Paper 1
Question 11
A circle with centre C has equation $x^2 + y^2 + 8x - 12y = 12$. (a) Find the coordinates of C and the radius of the circle. (b) The points P and Q lie on th... show full transcript
Worked Solution & Example Answer:A circle with centre C has equation $x^2 + y^2 + 8x - 12y = 12$ - AQA - A-Level Maths: Mechanics - Question 11 - 2017 - Paper 1
Step 1
Find the coordinates of C and the radius of the circle.
Answer
To find the coordinates of the center and the radius of the circle, we will rewrite the circle's equation in standard form. The standard form of a circle is given by:
where ( (h, k) ) are the coordinates of the center (C) and ( r ) is the radius.
Starting with the given equation:
We rearrange components:
- Group the x's and y's:
- Complete the square for the x's:
- Take half of 8 (which is 4) and square it (16), adding 16 to both sides:
- Complete the square for the y's:
- Take half of -12 (which is -6) and square it (36), adding 36 to both sides:
Thus, we find:
- The center C is ((-4, 6)) and the radius ( r = 8 ).
Step 2
Show that PQ has length \( \sqrt{3} \), where n is an integer.
Answer
To show that the length of chord PQ equals ( \sqrt{3} ), we first denote the coordinates of points P and Q lying on the circle. Since the origin (0,0) is the midpoint, we have:
Let P = ((x_1, y_1)) and Q = ((x_2, y_2)). Since the origin is the midpoint, we can set up the equations: [ x_1 + x_2 = 0 ] [ y_1 + y_2 = 0 ]
This implies ( x_1 = -x_2 ) and ( y_1 = -y_2 ).
Using the perpendicular distance from the center to the chord PQ: The radius is 8, and we will denote the distance from the center C to the chord as d: Using the Pythagorean theorem: [ r^2 = d^2 + (\frac{PQ}{2})^2 ] We know ( r = 8 ):
[ 64 = d^2 + (\frac{PQ}{2})^2 ]
Now focusing on d: The coordinates of C are ((-4, 6)). The distance d from C to the origin: [ d = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} ]
Substituting back into the equation: [ 64 = (2\sqrt{13})^2 + (\frac{PQ}{2})^2 ] [ 64 = 52 + (\frac{PQ}{2})^2 ] [ 12 = (\frac{PQ}{2})^2 ] [ PQ = 2\sqrt{12} = 4\sqrt{3} ] Thus we can express PQ in the required form, by letting ( n = 4 ): [ PQ = \sqrt{3} ] where n is an integer.