The circles $c_1$ and $c_2$ touch externally as shown. (a) Complete the following table: Circle Centre Radius Equation c_1 (-3, -2) 2 c_2 x^2 + y^2 - 2x - 2y - 7 = 0 (b) Find the co-ordinates of the point of contact of $c_1$ and $c_2$. (i) Hence, or otherwise, find the equation of the tangent, $t$, common to $c_1$ and $c_2$.

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The circles $c_1$ and $c_2$ touch externally as shown - Leaving Cert Mathematics - Question 4 - 2013

Question 4

The circles $c_1$ and $c_2$ touch externally as shown. (a) Complete the following table: Circle Centre Radius Equation c_1 (-3, -2) 2 c_2 x^2 + y^2 - 2x ... show full transcript

Worked Solution & Example Answer:The circles $c_1$ and $c_2$ touch externally as shown - Leaving Cert Mathematics - Question 4 - 2013

Step 1

Complete the following table:

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Answer

We need to deduce the missing values for circle $c_2$ .

Centre: For the equation $x^2 + y^2 - 2x - 2y - 7 = 0$ , we can rearrange this to

$(x-1)^2 + (y-1)^2 = 11$

Thus, the centre is $(1, 1)$ .

Radius: The radius is the square root of 11, so the radius is $oxed{3}$ .

The completed table is as follows:

Circle Centre Radius Equation c_1 (-3, -2) 2 c_2 (1, 1) 3 x^2 + y^2 - 2x - 2y - 7 = 0

Step 2

Find the co-ordinates of the point of contact of $c_1$ and $c_2$.

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Answer

To find the point of contact, we can divide the line segment joining the centres $(-3, -2)$ and $(1, 1)$ in the ratio of their radii, $2:3$ .

Using the section formula, the coordinates are calculated as follows:

$rac{(2 imes 1 + 3 imes -3)}{2 + 3}, rac{(2 imes 1 + 3 imes -2)}{2 + 3} = \left(\frac{(2 - 9)}{5}, \frac{(2 - 6)}{5}\right) = \left(-\frac{7}{5}, -\frac{4}{5}\right)$

Thus, the coordinates of the point of contact are $\left(-\frac{7}{5}, -\frac{4}{5}\right)$ .

Step 3

Hence, or otherwise, find the equation of the tangent, $t$, common to $c_1$ and $c_2$.

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Answer

To find the equation of the tangent, we first find the slope of the line joining the centres:

The slope from $(-3, -2)$ to $(1, 1)$ is:

$m_{centres} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-2)}{1 - (-3)} = \frac{3}{4}$

The tangent will be perpendicular to this line, hence the slope of the tangent $m$ will be:

$m = -\frac{4}{3}$

Using the point of contact $\left(-\frac{7}{5}, -\frac{4}{5}\right)$ and the slope in point-slope form, we have:

$y - \left(-\frac{4}{5}\right) = -\frac{4}{3}\left(x - \left(-\frac{7}{5}\right)\right)$

Rearranging gives:

$3y + 4x + 28/5 + 12/5 = 0 \implies 4x + 3y + 8 = 0$

Thus, the equation of the tangent $t$ is:

$4x + 3y + 8 = 0$

The circles $c_1$ and $c_2$ touch externally as shown - Leaving Cert Mathematics - Question 4 - 2013

Worked Solution & Example Answer:The circles $c_1$ and $c_2$ touch externally as shown - Leaving Cert Mathematics - Question 4 - 2013

Complete the following table:

Find the co-ordinates of the point of contact of $c_1$ and $c_2$.

Hence, or otherwise, find the equation of the tangent, $t$, common to $c_1$ and $c_2$.

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