If (a, 1/a),(b, 1/b),(c, 1/c),and (d, 1/d) ar | vedclass

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

Question

(C) Let the equation of the circle be $x^2 + y^2 = r^2$. Given the radius $r = 4$,the equation is $x^2 + y^2 = 16$.Since the points $(t, 1/t)$ lie on

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let the equation of the circle be $x^2 + y^2 = r^2$. Given the radius $r = 4$,the equation is $x^2 + y^2 = 16$.Since the points $(t, 1/t)$ lie on the circle,they must satisfy the equation:$t^2 + (1/t)^2 = 16$Multiplying by $t^2$,we get $t^4 - 16t^2 + 1 = 0$.This is a biquadratic equation in $t$ where the roots are $a, b, c, d$.The product of the roots of the equation $t^4 + 0t^3 - 16t^2 + 0t + 1 = 0$ is given by the constant term divided by the leading coefficient.Thus,$abcd = 1/1 = 1$.

If $(a, 1/a)$,$(b, 1/b)$,$(c, 1/c)$,and $(d, 1/d)$ are four distinct points on a circle of radius $4$ units,then $abcd$ is equal to

Similar Questions

$A$ circle having its centre at the origin passes through the three vertices of an equilateral triangle,the length of whose median is $9$ units. Then,the equation of that circle is

Let $S$ be a circle concentric with the circle $3x^2+3y^2+x+y-1=0$. If the length of the tangent drawn from a point $(2,-2)$ to the given circle is the radius of the circle $S$,then the power of the point $(2,1)$ with respect to the circle $S$ is

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

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

The equation of the circle which touches the lines $x = 0$,$y = 0$ and $3x + 4y = 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module