The limiting points of the co-axial system co | vedclass

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

Question

(C) The given circles are $S_1: x^2+y^2+2x-2y+2=0$ and $S_2: x^2+y^2-\frac{2}{5}x-\frac{16}{5}y+\frac{13}{5}=0$. The family of co-axial circles is giv

Vedclass · Accepted Answer

$(-1,1),\left(\frac{1}{5}, \frac{8}{5}\right)$

Vedclass · Answer

(C) The given circles are $S_1: x^2+y^2+2x-2y+2=0$ and $S_2: x^2+y^2-\frac{2}{5}x-\frac{16}{5}y+\frac{13}{5}=0$. The family of co-axial circles is given by $S_1 + \lambda(S_1 - S_2) = 0$ or $S_1 + k S_2 = 0$. The limiting points are the centers of the point circles in the system,which occur when the radius $r = 0$. For a circle $x^2+y^2+2gx+2fy+c=0$,the radius is $r = \sqrt{g^2+f^2-c}$. Setting $r=0$ gives $g^2+f^2=c$. The radical axis is $S_1 - S_2 = 0$ $\Rightarrow (2 + \frac{2}{5})x + (-2 + \frac{16}{5})y + (2 - \frac{13}{5}) = 0$ $\Rightarrow \frac{12}{5}x + \frac{6}{5}y - \frac{3}{5} = 0$ $\Rightarrow 4x + 2y - 1 = 0$. Any circle in the system is $S_1 + \lambda(4x + 2y - 1) = 0 \Rightarrow x^2+y^2+(2+4\lambda)x+(-2+2\lambda)y+(2-\lambda)=0$. For limiting points,$g^2+f^2=c \Rightarrow (1+2\lambda)^2 + (-1+\lambda)^2 = 2-\lambda$. $1+4\lambda+4\lambda^2 + 1-2\lambda+\lambda^2 = 2-\lambda \Rightarrow 5\lambda^2+3\lambda=0$. Thus,$\lambda=0$ or $\lambda=-\frac{3}{5}$. For $\lambda=0$,the center is $(-1, 1)$. For $\lambda=-\frac{3}{5}$,the center is $(-(1+2(-\frac{3}{5})), -(-1+(-\frac{3}{5}))) = (-1+\frac{6}{5}, 1+\frac{3}{5}) = (\frac{1}{5}, \frac{8}{5})$. The limiting points are $(-1, 1)$ and $(\frac{1}{5}, \frac{8}{5})$.

Components	$\%$ of the total cellular mass
$A$. Water	$1$. $10-15$
$B$. Proteins	$2$. $70-90$
$C$. Carbohydrates	$3$. $5-7$
$D$. Lipids	$4$. $3$
$E$. Nucleic acids	$5$. $2$
$F$. Ions	$6$. $1$

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

Similar Questions

The $9^{th}$ pair of cranial nerves in a frog is:

Match the following components of a cell with their respective percentage of the total cellular mass:

Components $\%$ of the total cellular mass

$A$. Water $1$. $10-15$

$B$. Proteins $2$. $70-90$

$C$. Carbohydrates $3$. $5-7$

$D$. Lipids $4$. $3$

$E$. Nucleic acids $5$. $2$

$F$. Ions $6$. $1$

The pulley and spring are massless,and friction is absent everywhere. The $5\,kg$ block is released from rest. The speed of the $5\,kg$ block when the $2\,kg$ block leaves contact with the ground is (take the force constant of the spring $K = 40\,N/m$ and $g = 10\,m/s^2$):

$A$ magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

Vedclass Test Series

Exam Paper Generator

Online Exam Module