Understanding the Quarter Circle and Inscribed Circle Geometry
The problem asks for the ratio between the radius (\(R\)) of a quarter circle and the radius (\(r\)) of a circle perfectly inscribed within that quarter circle. An inscribed circle is one that touches all sides of the shape it's inside. In this case, the inscribed circle touches the two straight sides (radii) of the quarter circle and the curved arc.
Visualizing and Setting Up the Geometry
Let's imagine the quarter circle positioned in the first quadrant of a coordinate plane, with its center at the origin \((0,0)\). The two straight sides lie along the positive x-axis and the positive y-axis. The radius of this quarter circle is \(R\).
Now, consider the circle inscribed within this quarter circle. Let its radius be \(r\). Since this circle touches both the x-axis and the y-axis, its center must be equidistant from both axes. Therefore, the coordinates of the center of the inscribed circle are \((r, r)\).
The inscribed circle also touches the arc of the quarter circle. This means the distance from the center of the quarter circle (the origin) to the center of the inscribed circle, plus the radius of the inscribed circle (\(r\)), must equal the radius of the quarter circle (\(R\)).
Calculating Distances and Setting Up the Equation
We need to find the distance between the center of the quarter circle \((0,0)\) and the center of the inscribed circle \((r, r)\). Using the distance formula, this distance is:
Distance \(= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Distance \(= \sqrt{(r - 0)^2 + (r - 0)^2}\) Distance \(= \sqrt{r^2 + r^2}\) Distance \(= \sqrt{2r^2}\) Distance \(= r\sqrt{2}\)
As established earlier, the radius of the quarter circle (\(R\)) is the sum of this distance and the radius of the inscribed circle (\(r\)):
\(R = (\text{Distance between centers}) + r\) \(R = r\sqrt{2} + r\)
Finding the Ratio R to r
We can factor out \(r\) from the right side of the equation:
\(R = r(\sqrt{2} + 1)\)
To find the ratio of \(R\) to \(r\), we rearrange the equation:
\(\frac{R}{r} = \frac{r(\sqrt{2} + 1)}{r}\) \(\frac{R}{r} = \sqrt{2} + 1\)
This can be written in the ratio format as \(R:r = (\sqrt{2} + 1):1\).
Conclusion
The ratio of the radius of the quarter circle (\(R\)) to the radius of the inscribed circle (\(r\)) is \((\sqrt{2} + 1):1\).
Similar Questions
- The chord AB of a circle with centre at O is \(2\sqrt{3}\) times the height of the minor segment. If P is the area of the sector OAB and Q is the area of the minor segment of the circle, then what is the approximate value of \(\frac{P}{Q}\)?
(Take \(\sqrt{3} = 1.7\) and \(\pi = 3.14\)) What is the area of the shaded region?
What is the ratio of the area of the shaded region to the area of the non-shaded region?
What is the radius of the circle with centre at $O_1$?
What is the radius of the circle with centre at $O_2$?
What is the sum of the areas of the two circles?
What is the area of the shaded region?
- What is the area of the region between two concentric circles, if the length of a chord of the outer circle touching the inner circle at a particular point of its circumference is 14 cm?
(Take \(\pi = \frac{22}{7}\)) What is the ratio of the area of the shaded region to that of the non-shaded region?
What is the perimeter of the shaded region?
Important Questions from Circles
The sum of the radius and diameter of a circle is 84 cm. What is the circumference of this circle?
The maximum area of a right-angled triangle inscribed in a circle of radius r is
The tangent at a point C of a circle and diameter AB when extended intersect at D, if ∠DCA = 110°, then ∠CBA is equal to
The equation of a circle with diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area of 154 sq. units is
The internal center of similitude of two circles $ (x - 1)^2 + (y - 3)^2 = 4 $ and $ (x + 5)^2 + (y - 9)^2 = 16 $ is
