Question

Find the equation of the curves for which the slope of tangent at any point is equal to the slope of the line connecting that point and centre of the circle $(x+1{)}^2+(y+1{)}^2=r^2$

• A. y+1 = m (x+1)
• B. y -1 = m (x-1)
• C. y = m(x-1)
• D. y-1 = mx

Answer 1

The correct option is A y+1 = m (x+1)

We want to find a curve for which slope of tangent at any point is equal to slope of the line connecting that point and centre of the circle $(x+1{)}^2+(y+1{)}^2=r^2$. Let the point on the curve be (x, y). Centre of the circle $(x+1{)}^2+(y+1{)}^2=r^{2}is(-1,-1)$ Slope of the line connecting (x, y) and (-1, -1) is $\frac{y+1}{x+1}$
Slope of tangent = slope of line connecting (x, y) and (-1, -1)
$\Rightarrow \frac{dy}{dx}=\frac{y+1}{x+1}\Rightarrow \frac{dy}{y+1}=\frac{dx}{x+1}\Rightarrow ln(y+1)=In(x+1)+c\Rightarrow ln\frac{(y+1)}{(x+1)}=c\Rightarrow y+1=e^c(x+1)$
Let the constant $e^c$ be equal to m
$\Rightarrow y+1=m(x+1)$