The centre of a circle lies on the line $x - 2y - 1 = 0$. The x-axis and the line $y = 6$ are tangents to the circle. Find the equation of this circle. A different circle has equation $x^2 + y^2 - 6x - 12y + 41 = 0$. Show that this circle and the circle in part (a) touch externally.

Leaving Cert Mathematics The Circle: The centre of a circle lies on t

The centre of a circle lies on the line $x - 2y - 1 = 0$ - Leaving Cert Mathematics - Question 4 - 2010

Question 4

The centre of a circle lies on the line $x - 2y - 1 = 0$. The x-axis and the line $y = 6$ are tangents to the circle. Find the equation of this circle. A different ... show full transcript

Worked Solution & Example Answer:The centre of a circle lies on the line $x - 2y - 1 = 0$ - Leaving Cert Mathematics - Question 4 - 2010

Step 1

The centre of a circle lies on the line $x - 2y - 1 = 0$.

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Answer

From the equation of the line, we can express $x$ in terms of $y$ :

$x = 2y + 1$

Let the center of the circle be $(h, k)$ . Since the x-axis is a tangent, the distance from the center to the x-axis must equal the radius $r$ , which gives:

$k = r$

The line $y = 6$ is another tangent, so:

$k + r = 6$

We can now establish two equations:

$k = r$
$k + r = 6$

By substituting Equation 1 into Equation 2, we find:

ightarrow 2r = 6 ightarrow r = 3$$ Thus, $k = 3$. Now substituting for $k$ into the expression for $h$ gives: $$h = 2(3) + 1 = 7$$ The center of the circle is at $(7, 3)$, and the radius $r = 3$. The equation of the circle is given by: $$(x - 7)^2 + (y - 3)^2 = 3^2$$ Expanding this results in: $$(x - 7)^2 + (y - 3)^2 = 9$$ This can also be written as: $$x^2 + y^2 - 14x - 6y + 49 = 0$$

Step 2

Show that this circle and the circle in part (a) touch externally.

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Answer

The second circle has the equation:

$x^2 + y^2 - 6x - 12y + 41 = 0.$
This can be rewritten to find its center and radius:

Complete the square:
- For $x$ : $x^2 - 6x = (x - 3)^2 - 9$
- For $y$ :
  $y^2 - 12y = (y - 6)^2 - 36$

Substituting these back into the equation gives:

$(x - 3)^2 - 9 + (y - 6)^2 - 36 + 41 = 0$

Simplifying leads to:

$(x - 3)^2 + (y - 6)^2 = 4,$
indicating that the center is at $(3, 6)$ and the radius is $r_2 = 2$ .

Now, we find the distance between the two centers:

Center of circle in (a): $(7, 3)$ .
Center of circle in (b): $(3, 6)$ .

The distance $d$ between the centers is given by:

ightarrow ext{. } ight]= 5$$ $$ ext{Sum of radii: } 3 + 2 = 5. \ ext{Thus, the circles touch externally.}$$

The centre of a circle lies on the line $x - 2y - 1 = 0$ - Leaving Cert Mathematics - Question 4 - 2010

Worked Solution & Example Answer:The centre of a circle lies on the line $x - 2y - 1 = 0$ - Leaving Cert Mathematics - Question 4 - 2010

The centre of a circle lies on the line $x - 2y - 1 = 0$.

Show that this circle and the circle in part (a) touch externally.

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