The number of common tangents to the circles | vedclass

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

Question

(D) The given circles are ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$.For the first circle ${x^2} + {y^2} - x = 0$,the centre ${C_1} = (\frac{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The given circles are ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$.For the first circle ${x^2} + {y^2} - x = 0$,the centre ${C_1} = (\frac{1}{2}, 0)$ and radius ${r_1} = \sqrt{(\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.For the second circle ${x^2} + {y^2} + x = 0$,the centre ${C_2} = (-\frac{1}{2}, 0)$ and radius ${r_2} = \sqrt{(-\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.The distance between the centres is ${C_1}{C_2} = \sqrt{(\frac{1}{2} - (-\frac{1}{2}))^2 + (0 - 0)^2} = \sqrt{1^2} = 1$.Since ${C_1}{C_2} = 1$ and ${r_1} + {r_2} = \frac{1}{2} + \frac{1}{2} = 1$,we have ${C_1}{C_2} = {r_1} + {r_2}$.This indicates that the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$.

The number of common tangents to the circles ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$ is:

Similar Questions

The area of an equilateral triangle inscribed in the circle $x^2+y^2-6x+2y-28=0$ is . . . . . . sq. units.

If $(3,-2)$ is the centre of the circle $S \equiv x^2+y^2+2gx+2fy-23=0$ and $A$ is a point on the circle $S=0$ such that its distance from a point $P(-1,-5)$ is least,then $A=$

The length of the intercept that the circle $x^2 + y^2 + 10x - 6y + 9 = 0$ makes on the $x$-axis is:

The line $x+y+1=0$ intersects the circle $x^2+y^2-4x+2y-4=0$ at the points $A$ and $B$. If $M(a, b)$ is the midpoint of $AB$,then $a-b=$

Observe the following statements:
$I$. The circle $x^2+y^2-6x-4y-7=0$ touches the $y$-axis.
$II$. The circle $x^2+y^2+6x+4y-7=0$ touches the $x$-axis.
Which of the following is a correct statement?

Vedclass Test Series

Exam Paper Generator

Online Exam Module