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The centre of a circle lies on the line $x + 2y - 6 = 0$ - Leaving Cert Mathematics - Question 4 - 2014

Question 4

The centre of a circle lies on the line $x + 2y - 6 = 0$. The x-axis and the y-axis are tangents to the circle. There are two circles that satisfy these conditions. ... show full transcript

Worked Solution & Example Answer:The centre of a circle lies on the line $x + 2y - 6 = 0$ - Leaving Cert Mathematics - Question 4 - 2014

Step 1

Case 1: $y = x$

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Answer

In this case, since the centre lies on the line, we substitute $y = x$ into the equation of the line:

$x + 2(x) - 6 = 0$

This simplifies to:

$3x - 6 = 0$

which gives:

$x = 2$

Thus, the coordinates of the centre of the circle are $(2, 2)$ . The radius $r$ is equal to the distance from the centre to either axis, which gives us:

$r = |2| = 2$

Therefore, the equation of the circle is:

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

This leads to:

$(x - 2)^2 + (y - 2)^2 = 4$

Step 2

Case 2: $y = -x$

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Answer

Now, we take the case where $y = -x$ . Again substituting this back into the equation of the line gives:

$x + 2(-x) - 6 = 0$

which simplifies to:

$-x - 6 = 0$

leading us to:

$x = -6$

Thus, the coordinates of the centre are $(-6, 6)$ . The radius in this scenario is:

$r = |6| = 6$

The equation of the circle in this case becomes:

$(x + 6)^2 + (y - 6)^2 = 6^2$

This yields:

$(x + 6)^2 + (y - 6)^2 = 36$

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