If the circles x^2+y^2-4x+6y+13-a^2=0 and x^2 | vedclass

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

Question

(D) The given equations of the circles are $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$. For the first circle,the center $C_1 = (2, -3)$ and the

Vedclass · Accepted Answer

none of these

Vedclass · Answer

(D) The given equations of the circles are $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$. For the first circle,the center $C_1 = (2, -3)$ and the radius $r_1 = \sqrt{2^2 + (-3)^2 - (13-a^2)} = \sqrt{4+9-13+a^2} = |a|$. For the second circle,the center $C_2 = (5, 1)$ and the radius $r_2 = \sqrt{5^2 + 1^2 - 17} = \sqrt{25+1-17} = \sqrt{9} = 3$. The distance between the centers $d = C_1C_2 = \sqrt{(5-2)^2 + (1-(-3))^2} = \sqrt{3^2 + 4^2} = 5$. For the circles to intersect at two distinct points,the condition is $|r_1 - r_2| Substituting the values,we get $||a| - 3| From $5  2$,which implies $a \in (-\infty, -2) \cup (2, \infty)$. From $||a| - 3| Combining both conditions,$a \in (-8, -2) \cup (2, 8)$.

If the circles $x^2+y^2-4x+6y+13-a^2=0$ and $x^2+y^2-10x-2y+17=0$ intersect in two distinct points,then '$a$' is

Similar Questions

If the circle $x^2+y^2-6x-12y+1=0$ cuts another circle $C$ orthogonally and the centre of the circle $C$ is $(-4, 2)$,then its radius is

The equation of the circle passing through the origin,having its center on the line $x + y = 4$,and intersecting the circle $x^2 + y^2 - 4x + 2y + 4 = 0$ orthogonally is:

If $(a, b)$ and $(c, d)$ are the internal and external centres of similitude of the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-6y+8=0$ respectively,then $(a+d)(b+c)=$

If the circles $x^2 + y^2 - 16x - 20y + 164 = r^2$ and $(x - 4)^2 + (y - 7)^2 = 36$ intersect at two distinct points,then

In the co-axial system of circles $x^2 + y^2 + 2gx + c = 0$,where $g$ is a parameter,if $c > 0$,then the circles are

Vedclass Test Series

Exam Paper Generator

Online Exam Module