A-Level Edexcel Maths Pure Radian Measure: The points $P(-3, 2)$, $Q(9, 10)

The points $P(-3, 2)$, $Q(9, 10)$ and $R(a, 4)$ lie on the circle $C_c$, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Question 7

The points $P(-3, 2)$, $Q(9, 10)$ and $R(a, 4)$ lie on the circle $C_c$, as shown in Figure 2. Given that $PR$ is a diameter of $C_c$; (a) show that $a = 13$; (b) ... show full transcript

Worked Solution & Example Answer:The points $P(-3, 2)$, $Q(9, 10)$ and $R(a, 4)$ lie on the circle $C_c$, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Step 1

show that a = 13

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Answer

To show that $a = 13$ , we need to find the slopes of the segments $PQ$ and $QR$ to confirm that they are perpendicular, since the segment $PR$ is a diameter of the circle.

Calculate the slope of PQ:

The formula for slope is given by: $m = \frac{y_2 - y_1}{x_2 - x_1}$

For points $P(-3, 2)$ and $Q(9, 10)$ : $m_{PQ} = \frac{10 - 2}{9 - (-3)} = \frac{8}{12} = \frac{2}{3}$
Calculate the slope of QR:

For points $Q(9, 10)$ and $R(a, 4)$ : $m_{QR} = \frac{4 - 10}{a - 9} = \frac{-6}{a - 9}$
Check for perpendicularity:

Since lines are perpendicular, the product of their slopes should be -1:

$m_{PQ} \cdot m_{QR} = -1$

Substituting the known values: $\frac{2}{3} \cdot \left(\frac{-6}{a - 9}\right) = -1$

This simplifies to: $\frac{-12}{3(a - 9)} = -1$

Solving for $a$ :

$12 = 3(a - 9)$
$12 = 3a - 27$
$3a = 39$
$a = 13$

Step 2

find an equation for C

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Answer

To find the equation of the circle $C$ , we can use the center and radius formula for a circle, which is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Where $(h, k)$ is the center and $r$ is the radius.

Determine the center of the circle:

The center, $O$ , lies at the midpoint of $P$ and $R$ . Thus, we find:

$h = \frac{-3 + 13}{2} = \frac{10}{2} = 5$ $k = \frac{2 + 4}{2} = \frac{6}{2} = 3$

Therefore, the center $O$ is at $(5, 3)$ .
Calculate the radius:

The radius can be calculated as the distance from the center $O(5, 3)$ to either point $P(-3, 2)$ or $R(13, 4)$ . Using point $P$ :

The formula for distance is given by: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Thus,

$r = \sqrt{(5 - (-3))^2 + (3 - 2)^2} = \sqrt{(5 + 3)^2 + (3 - 2)^2} = \sqrt{8^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65}$
Write the equation for C:

Substituting $h$ , $k$ , and $r$ into the standard form of the circle equation:

$(x - 5)^2 + (y - 3)^2 = 65$

The points $P(-3, 2)$, $Q(9, 10)$ and $R(a, 4)$ lie on the circle $C_c$, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Worked Solution & Example Answer:The points $P(-3, 2)$, $Q(9, 10)$ and $R(a, 4)$ lie on the circle $C_c$, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

show that a = 13

find an equation for C

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