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The circle k has centre C(1,−2) and chord [AB] where |AB| = 4√3 - Leaving Cert Mathematics - Question 3 - 2021

Question 3

The circle k has centre C(1,−2) and chord [AB] where |AB| = 4√3. The point D(3, 2) is the midpoint of the chord [AB], as shown below. Find the radius of k. Give ... show full transcript

Worked Solution & Example Answer:The circle k has centre C(1,−2) and chord [AB] where |AB| = 4√3 - Leaving Cert Mathematics - Question 3 - 2021

Step 1

Find the radius of k

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Answer

To find the radius of circle k, we need to use the triangle formed by the center C(1,−2), the midpoint D(3,2), and one of the endpoints of the chord A or B.

Calculate the distance |CD|:

$|CD| = \sqrt{(3 - 1)^2 + (2 - (-2))^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$
The length of the chord AB is given as |AB| = 4√3, and since D is the midpoint, we have:

$|AD| = |BD| = \frac{|AB|}{2} = \frac{4\sqrt{3}}{2} = 2\sqrt{3}$
Using the Pythagorean theorem in triangle CDB:

$r^2 = |CD|^2 + |AD|^2$

$r^2 = (2\sqrt{5})^2 + (2\sqrt{3})^2$

$r^2 = 20 + 12 = 32$

$r = \sqrt{32} = 4\sqrt{2}$

Thus, the radius of k is 4√2.

Step 2

Show that the circles c: x² + y² + 4x − 2y − 95 = 0 and s: (x − 7)² + (y − 13)² = 25 touch externally

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Answer

To show that the circles touch externally, we first find the centers and radii of both circles:

For circle c:
The equation can be rewritten as:
$x^2 + y^2 + 4x − 2y = 95$
Completing the square:

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

Thus, center C(-2, 1) and radius r₁ = 10.
For circle s:

The equation is in standard form:
$(x - 7)^2 + (y - 13)^2 = 5^2$
Thus, center S(7, 13) and radius r₂ = 5.
Calculate the distance between the centers C and S:

$d = \sqrt{((-2) - 7)^2 + (1 - 13)^2} = \sqrt{(-9)^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15$
Since r₁ + r₂ = 10 + 5 = 15, the circles touch externally.

Step 3

Find the coordinates of the centre of one of these circles, apart from circle s

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Answer

Let the center of the touching circle, called T, lie on the line passing through the centers of circles c and s.
The slope of line CS is:

$\text{slope} = \frac{13 - 1}{7 - (-2)} = \frac{12}{9} = \frac{4}{3}$

This means that lines from the center C and S create a relationship we can utilize for any point along this line. The equation of the line is:

$y - 1 = \frac{4}{3}(x + 2)$

Let’s use the known position of S(7, 13) as one endpoint and find another point on this line, ensuring it’s not coinciding with S:

Substitute $x = 4$ :

$y - 1 = \frac{4}{3}(4 + 2) = \frac{4}{3} imes 6 = 8 \implies y = 9$

A possible center coordinate for a touching circle, apart from circle s, is (4, 9).

The circle k has centre C(1,−2) and chord [AB] where |AB| = 4√3 - Leaving Cert Mathematics - Question 3 - 2021

Worked Solution & Example Answer:The circle k has centre C(1,−2) and chord [AB] where |AB| = 4√3 - Leaving Cert Mathematics - Question 3 - 2021

Find the radius of k

Show that the circles c: x² + y² + 4x − 2y − 95 = 0 and s: (x − 7)² + (y − 13)² = 25 touch externally

Find the coordinates of the centre of one of these circles, apart from circle s

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