A straight line x = y + 2 touches the circle 4(x 2+ y 2) = r 2. The value of r is
Finding the Radius of a Circle Tangent to a Line
The problem asks for the value of the radius \(r\) of a circle given by the equation \(4(x^2 + y^2) = r^2\) that is tangent to the straight line \(x = y + 2\).
Understanding the Equations
First, let's convert the given equations into their standard forms to easily identify the center and radius of the circle and the coefficients of the line.
The equation of the circle is \(4(x^2 + y^2) = r^2\). To get the standard form \( (x-h)^2 + (y-k)^2 = R^2 \), we divide by 4:
\(x^2 + y^2 = \frac{r^2}{4}\)
This equation represents a circle centered at the origin \((h, k) = (0, 0)\). The square of the radius is \(R^2 = \frac{r^2}{4}\), so the radius of this circle is \(R = \sqrt{\frac{r^2}{4}} = \frac{r}{2}\).
The equation of the straight line is \(x = y + 2\). To get the standard form \(Ax + By + C = 0\), we rearrange the terms:
\(x - y - 2 = 0\)
Comparing this to \(Ax + By + C = 0\), we have \(A=1\), \(B=-1\), and \(C=-2\).
Condition for Tangency: Distance from Center to Line
A fundamental concept in coordinate geometry is that a straight line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle.
The formula for the perpendicular distance \(d\) from a point \((h, k)\) to a line \(Ax + By + C = 0\) is given by:
\(d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}\)
In our case, the center of the circle is \((h, k) = (0, 0)\), and the line is \(x - y - 2 = 0\) with \(A=1\), \(B=-1\), \(C=-2\). The radius of the circle is \(R = \frac{r}{2}\).
The distance \(d\) from the center \((0, 0)\) to the line \(x - y - 2 = 0\) is:
\(d = \frac{|(1)(0) + (-1)(0) + (-2)|}{\sqrt{(1)^2 + (-1)^2}}\)
\(d = \frac{|0 + 0 - 2|}{\sqrt{1 + 1}}\)
\(d = \frac{|-2|}{\sqrt{2}}\)
\(d = \frac{2}{\sqrt{2}}\)
To rationalize the denominator, we can multiply the numerator and denominator by \(\sqrt{2}\):
\(d = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}\)
Equating Distance and Radius to Find r
Since the line touches (is tangent to) the circle, the distance from the center to the line must be equal to the radius of the circle.
\(d = R\)
We found \(d = \sqrt{2}\) and \(R = \frac{r}{2}\). So, we have:
\(\sqrt{2} = \frac{r}{2}\)
To find \(r\), we multiply both sides by 2:
\(r = 2 \times \sqrt{2}\)
\(r = 2\sqrt{2}\)
Thus, the value of \(r\) is \(2\sqrt{2}\).
| Geometric Element | Equation | Standard Form | Center / Coefficients | Radius |
|---|---|---|---|---|
| Circle | \(4(x^2 + y^2) = r^2\) | \(x^2 + y^2 = \frac{r^2}{4}\) | \((0, 0)\) | \(R = \frac{r}{2}\) |
| Line | \(x = y + 2\) | \(x - y - 2 = 0\) | \(A=1, B=-1, C=-2\) | N/A |
Summary of Steps
- Identify the center and radius (in terms of \(r\)) of the circle from its equation.
- Rewrite the line equation in the standard form \(Ax + By + C = 0\).
- Calculate the perpendicular distance from the circle's center to the line using the distance formula.
- Equate this distance to the radius of the circle.
- Solve the resulting equation for \(r\).
Following these steps, we found that \(r = 2\sqrt{2}\).
Revision Table: Circle and Line Tangency
| Concept | Description | Formula / Standard Form |
|---|---|---|
| Circle Equation | Standard form centered at \((h, k)\) with radius \(R\). | \((x-h)^2 + (y-k)^2 = R^2\) |
| Line Equation | Standard form. | \(Ax + By + C = 0\) |
| Distance from Point to Line | Perpendicular distance from \((x_1, y_1)\) to \(Ax + By + C = 0\). | \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\) |
| Tangency Condition | Line is tangent to circle if distance from center to line equals radius. | Distance \(d = R\) |
Additional Information: Geometric Interpretation
When a line touches a circle at exactly one point, it is called a tangent line. The radius drawn from the center of the circle to the point of tangency is always perpendicular to the tangent line. The distance calculation we performed \(\left(\frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}\right)\) finds the length of this perpendicular segment from the center \((h,k)\) to the line \(Ax+By+C=0\). For tangency, this length must be exactly equal to the radius \(R\).
In this specific problem, the circle is centered at the origin \((0,0)\). The line \(x-y-2=0\) has a positive slope (1). The distance from \((0,0)\) to this line being \(\sqrt{2}\) is a measure of how "far" the origin is from the line. Since this distance must be the radius \(\frac{r}{2}\), we solve for \(r\).
Important Questions from Circles
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