The equation of the circle having the lines y | vedclass

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

Question

(A) The given equation of the normals is $y^2 - 2y + 4x - 2xy = 0$.This can be factored as $y(y - 2) - 2x(y - 2) = 0$,which gives $(y - 2)(y - 2x) = 0

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 4y + 3 = 0$

Vedclass · Answer

(A) The given equation of the normals is $y^2 - 2y + 4x - 2xy = 0$.This can be factored as $y(y - 2) - 2x(y - 2) = 0$,which gives $(y - 2)(y - 2x) = 0$.Thus,the normals are the lines $y = 2$ and $y = 2x$.The intersection of these normals is the center $(h, k)$ of the circle.Solving $y = 2$ and $y = 2x$,we get $2 = 2x$,so $x = 1$.The center of the circle is $(1, 2)$.The general equation of a circle with center $(1, 2)$ is $(x - 1)^2 + (y - 2)^2 = r^2$,which simplifies to $x^2 + y^2 - 2x - 4y + 5 - r^2 = 0$.Since the circle passes through $(2, 1)$,we substitute these coordinates into the equation:$(2 - 1)^2 + (1 - 2)^2 = r^2$ $\Rightarrow 1^2 + (-1)^2 = r^2$ $\Rightarrow r^2 = 2$.Substituting $r^2 = 2$ into the circle equation: $x^2 + y^2 - 2x - 4y + 5 - 2 = 0$,which is $x^2 + y^2 - 2x - 4y + 3 = 0$.

The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals and passing through the point $(2, 1)$ is:

Similar Questions

The angle between the tangents drawn from the origin to the circle $(x - 7)^2 + (y + 1)^2 = 25$ is:

The line $x + y = 2$ is tangent to the curve $x^2 = 3 - 2y$ at its point

If the line $y = mx + c$ is a tangent to the circle $x^2 + y^2 = a^2$,then the point of contact is

Two tangents to the circle $x^{2}+y^{2}=4$ at the points $A$ and $B$ meet at $M(-4,0)$. The area of the quadrilateral $MAOB$, where $O$ is the origin, is

The equation of the tangents to the circle $x^2 + y^2 + 4x - 4y + 4 = 0$ which make equal intercepts on the positive coordinate axes is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module