The line $y = 3x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P(2, 2)$, as shown in Figure 1. The point $Q$ is the centre of $C$. (a) Find an equation of the straight line through $P$ and $Q$. Given that $Q$ lies on the line $y = 1$, (b) show that the $x$-coordinate of $Q$ is $5$. (c) find an equation for $C$.

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The line $y = 3x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P(2, 2)$, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

Question 9

The line $y = 3x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P(2, 2)$, as shown in Figure 1. The point $Q$ is the centre of $C$. (a) Find an eq... show full transcript

Worked Solution & Example Answer:The line $y = 3x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P(2, 2)$, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

Step 1

Find an equation of the straight line through P and Q.

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Answer

To find the equation of the line through points $P(2, 2)$ and $Q(x_Q, y_Q)$ , we first determine the gradient of the line $PQ$ .

The line $y = 3x - 4$ is tangent to circle $C$ at point $P(2, 2)$ . The gradient of this line is $3$ . The gradient of line $PQ$ should therefore be the negative reciprocal of $3$ , which is $- rac{1}{3}$ .

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$ Substituting for $P(2, 2)$ and the gradient $m = - rac{1}{3}$ :
$y - 2 = - rac{1}{3}(x - 2)$ Thus, we can rearrange this to obtain the equation: $3y - 6 = -x + 2$ or equivalently, $x + 3y = 8$ .

Step 2

Given that Q lies on the line y = 1, show that the x-coordinate of Q is 5.

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Answer

Since $Q$ lies on the line $y = 1$ , we replace $y$ in the equation obtained in part (a): $x + 3(1) = 8$ This simplifies to: $x + 3 = 8$ Therefore, we solve for $x$ : $x = 8 - 3 = 5.$ Thus, the $x$ -coordinate of $Q$ is indeed $5$ .

Step 3

find an equation for C.

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Answer

The center of the circle $C$ is at the point $Q(5, 1)$ . The radius of the circle can be calculated using the distance formula from the center $Q$ to the point of tangency $P(2, 2)$ : $r = ext{distance}(P, Q) = \\sqrt{(5 - 2)^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}.$ Using the standard form for the equation of a circle: $(x - h)^2 + (y - k)^2 = r^2,$ where $(h, k)$ is the center and $r$ is the radius, we plug in the values: $(x - 5)^2 + (y - 1)^2 = 10.$ This is the equation of circle $C$ .

The line $y = 3x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P(2, 2)$, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

Worked Solution & Example Answer:The line $y = 3x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P(2, 2)$, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

Find an equation of the straight line through P and Q.

Given that Q lies on the line y = 1, show that the x-coordinate of Q is 5.

find an equation for C.

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