The centre and radius of a circle x=4 aleft(f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given parametric equations are $x = 4a \left(\frac{1-t^2}{1+t^2}ight)$ and $y = \frac{8at}{1+t^2}$.Let $t = 	an 	heta$. Then $\cos 2	heta

Vedclass · Accepted Answer

$(0,0)$ and $4 a$ units

Vedclass · Answer

(B) The given parametric equations are $x = 4a \left(\frac{1-t^2}{1+t^2}ight)$ and $y = \frac{8at}{1+t^2}$.Let $t = 	an 	heta$. Then $\cos 2	heta = \frac{1-t^2}{1+t^2}$ and $\sin 2	heta = \frac{2t}{1+t^2}$.Substituting these,we get $x = 4a \cos 2	heta$ and $y = 4a \sin 2	heta$.Squaring and adding both equations:$x^2 + y^2 = (4a \cos 2	heta)^2 + (4a \sin 2	heta)^2$$x^2 + y^2 = 16a^2 (\cos^2 2	heta + \sin^2 2	heta)$$x^2 + y^2 = (4a)^2$.This represents a circle with centre $(0,0)$ and radius $4a$.

The centre and radius of a circle $x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right), y=\frac{8 a t}{1+t^{2}}$ are respectively:

Similar Questions

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of a circle with an area of $154$ square units. Find the equation of the circle.

The radius of the larger circle lying in the first quadrant and touching the line $4x + 3y - 12 = 0$ and the coordinate axes is:

$A$ circle of radius $5$ units touches the axes in the first quadrant. If the circle rolls along the $x-$axis in the positive $x-$direction for one complete revolution,find its equation in the new position.

The equations of the circle which pass through the origin and make intercepts of lengths $4$ and $8$ on the $x$ and $y$-axes respectively are

If $\alpha$ and $\beta$ are the roots of $x^2+2x-3=0$ and $\gamma$ and $\delta$ are the roots of $y^2-y-6=0$,then the equation of the circle having $(\alpha, \gamma)$ and $(\beta, \delta)$ as ends of a diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module