For the circle x-2=5 cos theta, y+1=5 sin the | vedclass

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

Question

(A) The given circle is $(x-2)^2 + (y+1)^2 = 25$,which has center $C(2, -1)$ and radius $R = 5$. The line is given by $x = 1 + \frac{r}{2}$ and $y = -

Vedclass · Accepted Answer

Chord of the circle other than diameter

Vedclass · Answer

(A) The given circle is $(x-2)^2 + (y+1)^2 = 25$,which has center $C(2, -1)$ and radius $R = 5$. The line is given by $x = 1 + \frac{r}{2}$ and $y = -2 + \frac{\sqrt{3}}{2} r$. Rearranging for $r$,we get $r = 2(x-1)$ and $r = \frac{2}{\sqrt{3}}(y+2)$. Equating these,$2(x-1) = \frac{2}{\sqrt{3}}(y+2) \implies \sqrt{3}x - y - (\sqrt{3}+2) = 0$. The perpendicular distance $d$ from the center $(2, -1)$ to the line is $d = \frac{|\sqrt{3}(2) - (-1) - (\sqrt{3}+2)|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{|2\sqrt{3} + 1 - \sqrt{3} - 2|}{\sqrt{3+1}} = \frac{|\sqrt{3}-1|}{2}$. Since $d Since the line does not pass through the center $(2, -1)$ (substituting $x=2, y=-1$ gives $\sqrt{3}(2) - (-1) - \sqrt{3} - 2 = \sqrt{3}-1 
eq 0$),it is a chord other than the diameter.

For the circle $x-2=5 \cos \theta, y+1=5 \sin \theta$ where $\theta$ is the parameter,the line $x=1+\frac{r}{2}, y=-2+\frac{\sqrt{3}}{2} r$ where $r$ is the parameter,is a

Similar Questions

$A$ rectangle $R$ with end points of one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2x - y + 4 = 0$,then the area of $R$ is

The radius of the circle passing through the points $(-1, 1)$,$(2, -1)$,and $(1, 0)$ is

The radical center of the three circles described with the three sides of a triangle as diameters is:

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. If $P(-5, 6)$ is a point such that $PA=PB=10$,then the equation of line $l$ is:

Let $A(2, 3), B(4, 5)$ be two points and let $C = (x, y)$ be a point such that $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$. If the area of $\Delta ABC = \sqrt{2}$ sq. unit,then the maximum number of positions of $C$ in the $xy$ plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module