The points of intersection of the line ax + b | vedclass

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

Question

(A) The line $ax + by = 0$ passes through the origin $(0, 0)$.Since $A(\alpha, 0)$ lies on the circle $x^2 + y^2 - 2x = 0$,substituting gives $\alpha^

Vedclass · Accepted Answer

$x^2 + y^2 + 5x + 5y + 12 = 0$

Vedclass · Answer

(A) The line $ax + by = 0$ passes through the origin $(0, 0)$.Since $A(\alpha, 0)$ lies on the circle $x^2 + y^2 - 2x = 0$,substituting gives $\alpha^2 - 2\alpha = 0$,so $\alpha = 0$ or $\alpha = 2$. If $\alpha = 0$,then $A = (0, 0)$.Since $B(1, \beta)$ lies on the circle,$1^2 + \beta^2 - 2(1) = 0$,so $\beta^2 = 1$,implying $\beta = 1$ or $\beta = -1$.Since the line $ax + by = 0$ passes through $(0, 0)$ and $(1, \beta)$,its equation is $y = \beta x$. Given $a 
eq b$,we find the intersection points are $(0, 0)$ and $(1, 1)$ (where $\beta=1$).The circle with diameter $AB$ where $A(0, 0)$ and $B(1, 1)$ is $(x - 0)(x - 1) + (y - 0)(y - 1) = 0$,which simplifies to $x^2 + y^2 - x - y = 0$.The center is $(1/2, 1/2)$ and radius is $r = \sqrt{(1/2)^2 + (1/2)^2} = 1/\sqrt{2}$.The image of the center $(1/2, 1/2)$ in the line $x + y + 2 = 0$ is $(h, k)$ such that $\frac{h - 1/2}{1} = \frac{k - 1/2}{1} = -2 \frac{1/2 + 1/2 + 2}{1^2 + 1^2} = -3$. Thus $h = -2.5, k = -2.5$.The image circle is $(x + 2.5)^2 + (y + 2.5)^2 = (1/\sqrt{2})^2$,which simplifies to $x^2 + y^2 + 5x + 5y + 12 = 0$.

The points of intersection of the line $ax + by = 0$ $(a \neq b)$ and the circle $x^2 + y^2 - 2x = 0$ are $A(\alpha, 0)$ and $B(1, \beta)$. The image of the circle with $AB$ as a diameter in the line $x + y + 2 = 0$ is:

Similar Questions

The maximum area of a rectangle inscribed in the circle $(x+1)^{2}+(y-3)^{2}=64$ is

If the midpoint of the chord intercepted by the circle $x^2+y^2-8x+10y+5=0$ on the line $2x+y+2=0$ is $(h, k)$,then $k+4h=$

If the radical axis of the circles $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ forms a triangle of area $A$ with the coordinate axes,then the value of $\frac{1}{A}$ is

The area of the circle centered at $(1, 2)$ and passing through $(4, 6)$ is

The image of the point $(3, 4)$ with respect to the radical axis of the circles $x^2 + y^2 + 8x + 2y + 10 = 0$ and $x^2 + y^2 + 7x + 3y + 10 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module