# Line normal to a circle

1. May 9, 2013

### Saitama

1. The problem statement, all variables and given/known data
A variable line ax+by+c=0, where a,b,c are in A.P (arithmetic progression), is normal to a circle $(x-\alpha)^2+(y-\beta)^2=\gamma$, which is orthogonal to circle $x^2+y^2-4x-4y-1=0$. The value of $\alpha+\beta+\gamma$ is equal to
A)3
B)5
C)10
D)7

2. Relevant equations
Condition for orthogonality of two circle: $2g_1g_2+2f_1f_2=c_1+c_2$

3. The attempt at a solution
From the above condition for orthogonality: $4\alpha+4\beta=\alpha^2+\beta^2-\gamma-1$

Since the line is normal to the circle, it passes through the centre of circle i.e. $a\alpha+b\beta+c=0$. What should I do next?

I have 2b=a+c but where should I use this?

Any help is appreciated. Thanks!

2. May 9, 2013

Anyone?

3. May 9, 2013

### Staff: Mentor

I don't see that you have used the given information that a, b, and c are in arithmetic progression.

4. May 9, 2013

### Saitama

I know and that's my question. Where am I supposed to use this? Replace c with 2b-a?

5. May 9, 2013

### Staff: Mentor

Assuming the common difference to be d, you have b = a + d, and c = a + 2d. This makes your line equation
ax + (a + d) y + a + 2d = 0.

Solve for y to get y = -a/(a + d) * x + (-a - 2d)/(a + d). This form gives you the slope of the line very easily.

6. May 9, 2013

### Saitama

I am still a bit lost. Why do we even need the slope here?

7. May 10, 2013

### haruspex

I don't believe there's enough information here. You have 3 equations and 6 unknowns. Now, it could just happen that the one variable of interest is determinable, but I don't see it. Substitute for γ using γ = Z - α-β, substitute for c using c = 2b-a. That gets you to two equations and 5 unknowns. You can use one equation to eliminate one more unknown, and hope that some huge magic cancellation occurs to eliminate 3 more (all except Z), but it doesn't.

8. May 10, 2013

### Saitama

This is from my test paper and I don't think that it requires too much time solving questions. But there is no guarantee that the question is correct.
I have the solution booklet too, I don't see from where did they get the values of $\alpha$ and $\beta$.

Solution:
(A part of the solution is on the different page so I am writing down the equations from the first page and attaching the rest of the steps.)
$ax+by+c=0$
$ax-2b+c=0$
$(1,-2)$ is the centre of cricle

#### Attached Files:

• ###### steps.jpg
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9. May 10, 2013

### haruspex

Based on that, I think I've figured it out. The question ought to read:
I suppose that's what they meant by 'variable' line, but it should have been made much clearer.
All such lines pass through the point (1, -2),

10. May 10, 2013

### Saitama

That means I can assume any line and continue with that?

How did the solution even come up with the values of $\alpha$ and $\beta$?

11. May 10, 2013

### haruspex

No - ALL lines of the form ax+by+c=0, where a,b,c are in arithmetic progression. That fixes where the centre of the circle must be; it must be a point that all such lines go through.

12. May 11, 2013

### Saitama

Thanks haruspex!

I assumed two equations, i.e $x+2y+3=0$ and $y+2=0$, this gave me $\alpha$ and $\beta$. I substituted them in the equation for orthogonality and found gamma.

Thank you! :)