If the line y = x + 3 intersects the circle x | vedclass

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

Question

(A) The equation of any circle passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x - y + 3 = 0$ is given by $S

Vedclass · Accepted Answer

$x^2 + y^2 + 3x - 3y - a^2 + 9 = 0$

Vedclass · Answer

(A) The equation of any circle passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x - y + 3 = 0$ is given by $S + \lambda L = 0$.$x^2 + y^2 - a^2 + \lambda(x - y + 3) = 0$$x^2 + y^2 + \lambda x - \lambda y + (3\lambda - a^2) = 0$The center of this circle is $\left(-\frac{\lambda}{2}, \frac{\lambda}{2}\right)$.Since $AB$ is the diameter,the center must lie on the line $y = x + 3$.Substituting the center coordinates into the line equation: $\frac{\lambda}{2} = -\frac{\lambda}{2} + 3$.$\lambda = 3$.Substituting $\lambda = 3$ into the circle equation: $x^2 + y^2 + 3x - 3y + (3(3) - a^2) = 0$.$x^2 + y^2 + 3x - 3y + 9 - a^2 = 0$.

If the line $y = x + 3$ intersects the circle $x^2 + y^2 = a^2$ at points $A$ and $B$,then the equation of the circle with $AB$ as its diameter is:

Similar Questions

Let $a=1+i$ and $z=x+iy$. If the curve $z\bar{z}+az+\bar{a}\bar{z}-4=0$ is cut by the straight line $(z+\bar{z})-i(z-\bar{z})+2=0$ at two points $A$ and $B$, then the equation of the circle passing through the origin, $A$ and $B$ is

If $(a, b)$ is the common point for the circles $x^2+y^2-4x+4y-1=0$ and $x^2+y^2+2x-4y+1=0$,then $a^2+b^2=$

If the circles $x^2 + y^2 + 2ax + c = 0$ and $x^2 + y^2 + 2by + c = 0$ touch each other,then:

The point of contact of the given circles $x^2 + y^2 - 6x - 6y + 10 = 0$ and $x^2 + y^2 = 2$ is

Let $x+y=0$ be the radical axis of the circles $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. If the radius of the circle $S=0$ is $1$,then find the value of $g+f$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module