Photo AI
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 $x - 2y - 1 = 0$.
Answer
From the equation of the line, we can express in terms of :
Let the center of the circle be . Since the x-axis is a tangent, the distance from the center to the x-axis must equal the radius , which gives:
The line is another tangent, so:
We can now establish two equations:
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$$Step 2
Show that this circle and the circle in part (a) touch externally.
Answer
The second circle has the equation:
This can be rewritten to find its center and radius:
- Complete the square:
- For :
- For :
Substituting these back into the equation gives:
Simplifying leads to:
indicating that the center is at and the radius is .
Now, we find the distance between the two centers:
- Center of circle in (a): .
- Center of circle in (b): .
The distance between the centers is given by:
ightarrow ext{. } ight]= 5$$ $$ ext{Sum of radii: } 3 + 2 = 5. \ ext{Thus, the circles touch externally.}$$