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 $y = 1$, 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 8 - 2006 - Paper 2

Question 8

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 8 - 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 $Q(x_Q, 1)$ , we first determine the gradient of the line $PQ$ . The gradient is given by:

$ext{Gradient} = rac{y_Q - y_P}{x_Q - x_P} = rac{1 - 2}{x_Q - 2} = rac{-1}{x_Q - 2}$

Since the line $y = 3x - 4$ is a tangent to the circle at point $P$ , the gradient of this tangent line is $3$ . Therefore, the slopes must be negative reciprocals:

$3 imes rac{-1}{x_Q - 2} = -1$

Solving for $x_Q$ gives:

$3 = x_Q - 2 \\ x_Q = 5$

Thus, the x-coordinate of $Q$ is $5$ . We can now use the point-slope form of the equation of the line:

$y - 2 = rac{-1}{3}(x - 2)$

Therefore, the equation becomes:

$y = - rac{1}{3}x + rac{8}{3}$ .

Step 2

show that the x-coordinate of Q is 5.

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Answer

We have already established that the y-coordinate of $Q$ is $1$ . Using the line equation, we substitute:

$1 = 3x - 4 \\ 3x = 5 \\ x = 5.$

This confirms that the $x$ -coordinate of $Q$ is indeed $5$ .

Step 3

Find an equation for C.

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Answer

The circle $C$ has a center at $Q(5, 1)$ and passes through point $P(2, 2)$ . The radius $r$ can be calculated using the distance formula:

$r = ext{distance}(Q, P) = ext{sqrt}((5 - 2)^2 + (1 - 2)^2) = ext{sqrt}(9 + 1) = ext{sqrt}(10).$

The general equation for a circle is given by:

$(x - h)^2 + (y - k)^2 = r^2,$ where $(h, k)$ is the center.

Substituting $Q(5, 1)$ and $r^2 = 10$ , we arrive at:

$(x - 5)^2 + (y - 1)^2 = 10.$

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 8 - 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 8 - 2006 - Paper 2

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

show that the x-coordinate of Q is 5.

Find an equation for C.

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