Question

The differential equation of all straight lines touching the circle \(x^2+y^2=a^2\) is

• A. \(\left(y-\frac{dy}{dx}\right)^2=a^2\left[1+\left(\frac{dy}{dx}\right)^2\right]\)
• B. \(\left(y-x\frac{dy}{dx}\right)^2=a^2\left[1+\left(\frac{dy}{dx}\right)^2\right]\)
• C. \(\left(y-x\frac{dy}{dx}\right)=a^2\left[1+\left(\frac{dy}{dx}\right)\right]\)
• D. \(\left(y-\frac{dy}{dx}\right)=a^2\left[1-\frac{dy}{dx}\right]\)

Hint:

For family of tangents to \(x^2+y^2=a^2\), write \(y=mx+c\), apply tangency condition \(c^2=a^2(1+m^2)\), then replace \(m=\frac{dy}{dx}\).

Answer 1

The Correct Option is B

Step 1: Write equation of a straight line with slope form.
General line:
\[ y=mx+c \]
Step 2: Condition for tangency with circle \(x^2+y^2=a^2\).
Distance of line from origin must be \(a\):
\[ \frac{|c|}{\sqrt{1+m^2}}=a \Rightarrow c^2=a^2(1+m^2) \]
Step 3: Express \(c\) in terms of \(x,y,m\).
From line:
\[ c=y-mx \]
Step 4: Substitute into tangency condition.
\[ (y-mx)^2=a^2(1+m^2) \]
Step 5: Replace \(m\) by \(\frac{dy}{dx}\).
Since \(m\) is slope:
\[ m=\frac{dy}{dx} \]
So differential equation becomes:
\[ \left(y-x\frac{dy}{dx}\right)^2 =a^2\left[1+\left(\frac{dy}{dx}\right)^2\right] \]
Final Answer:
\[ \boxed{\left(y-x\frac{dy}{dx}\right)^2=a^2\left[1+\left(\frac{dy}{dx}\right)^2\right]} \]