If a circle C, whose radius is 3, touch | vedclass

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

Question

Given circle is:

${x^2} + {y^2} + 2x - 4y - 4 = 0$

$	herefore $ it center is $\left( { - 1,2} ight)$ and radius is $3$ units.

Let $A=

Vedclass · Accepted Answer

$2\sqrt 5$

Vedclass · Answer

Given circle is:

${x^2} + {y^2} + 2x - 4y - 4 = 0$

$	herefore $ it center is $\left( { - 1,2} ight)$ and radius is $3$ units.

Let $A=(x,y)$ be the center of the circle $C$ 

$	herefore \frac{{x - 1}}{2} = 2 \Rightarrow x = 5$

$ \Rightarrow y = 2$

So the center of $C$ is $(5,2)$ and its radius is $3$

$	herefore $ equation of center $C$ is:

${x^2} + {y^2} - 10x - 4y + 20 = 0$

$	herefore $ The length of the intercept it cuts on the $X$-axis

$ = 2\sqrt {{g^2} - c}  = 2\sqrt {25 - 20}  = 2\sqrt 5 $

If a circle $C,$ whose radius is $3,$ touches externally the circle, $x^2 + y^2 + 2x - 4y - 4 = 0$ at the point $(2, 2),$ then the length of the intercept cut by circle $c,$ on the $x-$ axis is equal to

Similar Questions

A circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is

The circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ bisects the circumference of the circle ${x^2} + {y^2} + 2g'x + 2f'y + c' = 0$, if

If the line $x\, cos \theta + y\, sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} \,x - 6y + 20 = 0$, then the value of $\theta$ is :

The equation of the circle which passes through the point of intersection of circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and having its centre on $y$ - axis, will be

Two tangents are drawn from a point $P$ on radical axis to the two circles touching at $Q$ and $R$ respectively then triangle formed by joining $PQR$ is