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If the circle $x^2+y^2+4x+22y+l=0$ bisects the circumference of the circle $x^2+y^2-2x+8y-m=0$,then $l+m$ is equal to

(A)$\ 60$

(B)$\ 50$

(C)$\ 46$

(D)$\ 40$

I don't know the condition when one circle intersects the circumference of other circle. So could not solve this question. Can someone help me in this question?

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  • $\begingroup$ @thanasissdr,using this condition,i could not find $l+m$. $\endgroup$
    – diya
    Jul 31, 2015 at 7:10
  • $\begingroup$ The two circles must intersect at two points, where the two points of intersection must form a diagonal of the second circle. Start with the points of intersection: points will be common to both circles, so $x^2+y^2+4x+22y+l = x^2+y^2-2x+8y-m$. This will give a straight line, which depends on $l+m$. $\endgroup$
    – Maciek
    Jul 31, 2015 at 7:19

3 Answers 3

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Circle $C_1: x^2+y^2+4x+22y+l=0$ has its center $(-2, -11)$ & a radius $\sqrt{(-2)^2+(-11)^2-l}=\sqrt{125-l}$

Similarly, circle $C_2: x^2+y^2-2x+8y-m=0$ has its center $(1, -4)$ & a radius $\sqrt{(1)^2+(-4)^2-(-m)}=\sqrt{m+17}$

Now, solving the equations of circles $C_1$ & $C_2$ by substituting the value of $(x^2+y^2)$ from $C_2$ into $C_1$, we get the $\color{blue}{\text{equation of common chord}}$ as follows $$(2x-8y+m)+4x+22y+l=0$$ $$\color{blue}{6x+14y+(l+m)=0}\tag 1$$

Now, since the circumference of circle $C_2$ is bisected by the circle $C_1$ hence the center $(1, -4)$ of circle $C_2$ must lie on the common chord or in other words, the common chord must pass through the center of circle $C_2$

Now, satisfying the above equation of common chord by center point $(1, -4)$ as follows $$6(1)+14(-4)+(l+m)=0$$ $$6-56+l+m=0$$ $$\bbox[5px, border:2px solid #C0A000]{l+m=50}$$

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The condition for a circle (Circle A) to bisect the circumference of another circle (Circle B) is:

The common chord of A and B should pass through the center of B

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  • $\begingroup$ Welcome to Math.SE! This does not appear to provide a complete answer to the question and might be better left as a comment below the post - you can always comment on your own posts, and once you have sufficient reputation, you can comment on any post. $\endgroup$ May 1, 2019 at 3:13
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Let us consider the equation of bisecting circle to be "S1"...and the equation of bisected circle be "S2"

S1: x²+y² +4x+22y+l

S2: x²+y²-2x+8y-m

In order to calculate the value of "l+m"......we first need to calculate the value of common tangent...... Let us denote the eqn of common tangent with "L"

The eqn of common tangent will be = S1-S2

So eqn of common tangent is= x²+y²+4x+22y+l - (x²+y²-2x+8y-m)

Eqn of common tangent = 6x+14y+(l+m)=0

Also the centre of the bisected circle should lie on the common tangent.....

Centre of bisected circle = (1,-4). (-g,-f)

Putting the coordinates of centre of the circle in eqn of common tangent.....

6(1)+14(-4)+(l+m)=0

6-56+(l+m)=0

-50 +(l+m)=0

l+m=50

Hence the correct option is (B)

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  • $\begingroup$ I think you meant to say common chord or radical axis instead of common tangent. $\endgroup$ Dec 29, 2016 at 9:50

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