The lines represented by 5x^2-xy-5x+y=0 are n | vedclass

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

Question

(A) The given circle is $S^{\prime} \equiv x^2+y^2-2x+2y-7=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=1, c=-7$. The centre is $C_1(-g, -

Vedclass · Accepted Answer

$2y-7=0$

Vedclass · Answer

(A) The given circle is $S^{\prime} \equiv x^2+y^2-2x+2y-7=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=1, c=-7$. The centre is $C_1(-g, -f) = (1, -1)$ and the radius is $r_1 = \sqrt{g^2+f^2-c} = \sqrt{1+1+7} = 3$. The pair of lines $5x^2-xy-5x+y=0$ are normals to the circle $S=0$. Factoring the equation: $x(5x-y) - 1(5x-y) = 0 \Rightarrow (x-1)(5x-y) = 0$. The lines are $x=1$ and $y=5x$. The intersection of these normals is the centre $C_2$ of circle $S=0$. Solving $x=1$ and $y=5x$,we get $x=1, y=5$. So,$C_2(1, 5)$. Since the circles touch externally,the distance between centres $C_1C_2 = r_1 + r_2$. $C_1C_2 = \sqrt{(1-1)^2 + (5 - (-1))^2} = \sqrt{0^2 + 6^2} = 6$. Thus,$6 = 3 + r_2 \Rightarrow r_2 = 3$. The equation of circle $S=0$ with centre $(1, 5)$ and radius $3$ is $(x-1)^2 + (y-5)^2 = 3^2$. $x^2 - 2x + 1 + y^2 - 10y + 25 = 9 \Rightarrow x^2 + y^2 - 2x - 10y + 17 = 0$. The chord of contact of point $C_1(1, -1)$ with respect to $S=0$ is given by $T=0$. $x(1) + y(-1) - 1(x+1) - 5(y-1) + 17 = 0$. $x - y - x - 1 - 5y + 5 + 17 = 0$. $-6y + 21 = 0$ $\Rightarrow 6y = 21$ $\Rightarrow 2y = 7$ $\Rightarrow 2y-7=0$.

The lines represented by $5x^2-xy-5x+y=0$ are normals to a circle $S=0$. If this circle touches the circle $S^{\prime} \equiv x^2+y^2-2x+2y-7=0$ externally,then the equation of the chord of contact of the centre of $S^{\prime}=0$ with respect to $S=0$ is

Similar Questions

The circles $x^2 + y^2 = 9$ and $x^2 + y^2 - 12y + 27 = 0$ touch each other. The equation of their common tangent is

Find the condition that the pair of lines joining the origin to the points of intersection of the line $y = mx + c$ and the circle $x^{2} + y^{2} = a^{2}$ are at right angles to each other.

If the point of intersection of the tangents drawn at the points where the line $5x + y + 1 = 0$ cuts the circle $x^2 + y^2 - 2x - 6y - 8 = 0$ is $(a, b)$,then $5a + b =$

If the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2-4x-4y+8=0$ meets the circle in two distinct points and it also makes an angle $45^{\circ}$ with the positive $X$-axis in the positive direction,then $(h, k)$ cannot be

The combined equation of the direct common tangents of the circles $x^2+y^2-2x-2y-2=0$ and $x^2+y^2+4x+6y+12=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module