Question:
Find the differential equation of all circles in the first quadrant which touch both the coordinate axes.
Find the differential equation of all circles in the first quadrant which touch both the coordinate axes.
Show Hint
Use implicit differentiation to find the slope of tangents and eliminate parameters by substitution.
Updated On: June 02, 2025
Hide Solution
Verified By Collegedunia
Solution and Explanation
Step 1: The general equation of a circle touching both axes is: \[ (x - r)^2 + (y - r)^2 = r^2, \] where \( r \) is the radius of the circle, and the center is at \( (r, r) \).
Step 2: Differentiate implicitly with respect to \( x \): \[ 2(x - r) + 2(y - r) \frac{dy}{dx} = 0 \quad \Rightarrow \quad (x - r) + (y - r) \frac{dy}{dx} = 0. \]
Step 3: Eliminate \( r \) using the relationship \( x^2 + y^2 = 2xr \): Substitute \( r = \frac{x^2 + y^2}{2x} \) into the equation: \[ \frac{dy}{dx} = -\frac{x - \frac{x^2 + y^2}{2x}}{y - \frac{x^2 + y^2}{2x}}. \]
Was this answer helpful?
0
0
Top Questions on Differentiation
- If \( x = e^{\frac{x}{y}} \), then prove that \( \frac{dy}{dx} = \frac{x - y}{x \log x} \).
- CBSE CLASS XII - 2025
- Mathematics
- Differentiation
- The area (in sq. units) of the region given by \( R = \left\{ (x, y) : \dfrac{y^2}{2} \leq x \leq y + 4 \right\} \) is
- AP EAPCET - 2025
- Mathematics
- Differentiation
- Evaluate \[ \lim_{n \to \infty} \frac{1}{2n} \left( \sin \frac{\pi}{2n} + \sin \frac{\pi}{n} + \sin \frac{2\pi}{2n} + \dots \right) = ? \]
- AP EAPCET - 2025
- Mathematics
- Differentiation
- Evaluate the integral: \[ I = \int_0^x \frac{t^2}{\sqrt{a^2 + t^2}} dt \]
- AP EAPCET - 2025
- Mathematics
- Differentiation
- If \( k \in N \) then \( \lim_{n\to\infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{kn} \right] = \) (Note: The last term should be \( \frac{1}{n+ (k-1)n} = \frac{1}{kn} \) or sum up to \(n+(k-1)n\). The given form \(1/kn\) as the endpoint of the sum means sum from \(r=1\) to \((k-1)n\). The sum is usually \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). If the last term is \( \frac{1}{kn} \), it means \( n+r = kn \implies r = (k-1)n \). So it's \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \).) Let's assume the sum goes up to \( \frac{1}{n+(k-1)n} = \frac{1}{kn} \). So the sum is \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). No, this seems to be \( \frac{1}{n+1} + \dots + \frac{1}{n+(kn-n)} \). The sum should be written as \( \sum_{i=1}^{(k-1)n} \frac{1}{n+i} \). The dots imply the denominator goes up. The last term is \( \frac{1}{kn} \). This means the sum is actually \( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{n+(k-1)n} \). The number of terms is \( (k-1)n \).
- AP EAPCET - 2025
- Mathematics
- Differentiation
View More Questions
Questions Asked in CBSE CLASS XII exam
- If \( p \) and \( q \) are respectively the order and degree of the differential equation \( \frac{d}{dx} \left( \frac{dy}{dx} \right)^3 = 0 \), then \( (p - q) \) is:
- CBSE CLASS XII - 2025
- Differential Equations
- Show the refraction of light wave at a plane interface using Huygens' principle and prove Snell's law.
- CBSE CLASS XII - 2025
- Wave Optics
- The threshold voltage of a silicon diode is 0.7 V. It is operated at this point by connecting the diode in series with a battery of \( V \) volt and a resistor of 1000 \( \Omega \). Find the value of \( V \) when the current drawn is 15 mA.
- CBSE CLASS XII - 2025
- Semiconductor electronics: materials, devices and simple circuits
- Draw the structure of the major monohalo product for each of the following reactions:
- CBSE CLASS XII - 2025
- Haloalkanes And Haloarenes
- If a line makes angles of \( \frac{3\pi}{4} \) and \( \frac{\pi}{3} \) with the positive directions of \( x \), \( y \), and \( z \)-axes respectively, then \( \theta \) is:
View More Questions
