If the circles x^2 + y^2 - 2ax + c = 0 and x^ | vedclass

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

Question

(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 - 2ax + c = 0$,we have $g_1 = -a$,$f_1 = 0$,an

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 - 2ax + c = 0$,we have $g_1 = -a$,$f_1 = 0$,and $c_1 = c$.For the second circle $x^2 + y^2 + 2by + 2\lambda = 0$,we have $g_2 = 0$,$f_2 = b$,and $c_2 = 2\lambda$.The condition for two circles to intersect orthogonally is $2(g_1g_2 + f_1f_2) = c_1 + c_2$.Substituting the values,we get $2((-a)(0) + (0)(b)) = c + 2\lambda$.$0 = c + 2\lambda$.Therefore,$2\lambda = -c$,which implies $\lambda = -\frac{c}{2}$.Since $-\frac{c}{2}$ is not among the options $A, B, C$,the correct choice is $D$.

If the circles $x^2 + y^2 - 2ax + c = 0$ and $x^2 + y^2 + 2by + 2\lambda = 0$ intersect orthogonally,then the value of $\lambda$ is

Similar Questions

$A$ circle passes through $(0, 0)$ and $(1, 0)$ and touches the circle ${x^2} + {y^2} = 9$. Find the centre of the circle.

The radical axis of circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

If $(p, q)$ is the centre of the circle which cuts the three circles $x^2+y^2-2x-4y+4=0$,$x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y-11=0$ orthogonally,then $p+q=$

Find the equation of a circle with center $(4, 3)$ that touches the circle $x^2 + y^2 = 1$ internally.

Vedclass Test Series

Exam Paper Generator

Online Exam Module