Question

Circle with centre $(a, b)$ is drawn to cut two circles $x^2+y^2+6x+5=0$ & $x^2+y^2-6y+5=0$ orthogonally, then:

• A. $\frac{a-b}{a}=2$ (where $a\neq 0$)
• B. $\frac{a-b}{a}=3$ (where $a\neq 0$)
• C. $(a, b)$ lies on line $x+y=0$
• D. The circle always passes through two fixed points $(-1, 2)$ and $(-2, 1)$

Answer 1

Answer: (A), (C), (D)

Options 1, 3, and 4 are correct.

Answer 2

Solution:

1. Write the equations of the given circles in center‐radius form:

   • $x^2+y^2+6x+5=0$

     → $(x+3)^2+y^2=4$ (center $(-3,0)$, radius 2)

   • $x^2+y^2-6y+5=0$

     → $x^2+(y-3)^2=4$ (center $(0,3)$, radius 2)

2. Let the circle with center $(a,b)$ (and radius $r$) be orthogonal to both. The orthogonality condition is:

   $$\text{(distance between centers)}^2 = r^2 + (\text{given circle's radius})^2.$$

   For the first circle:

   $$(a+3)^2 + b^2 = r^2+4 \quad \text{(1)}$$

   For the second circle:

   $$a^2+(b-3)^2 = r^2+4 \quad \text{(2)}$$

   Subtract (1) from (2):

   $$[a^2+(b-3)^2] - [(a+3)^2+b^2] = 0.$$

   Expanding and simplifying:

   $$a^2+(b^2 - 6b +9) - (a^2+6a+9+b^2)= -6b-6a = 0,$$

   which gives:

   $$a+b=0 \quad \text{or} \quad b=-a.$$

   Thus, the center $(a,b)$ lies on the line $x+y=0$ [Option (3)].

3. Now, substituting $b = -a$ into the expression $\frac{a-b}{a}$:

   $$\frac{a - (-a)}{a} = \frac{2a}{a} = 2.$$

   So, we get $\frac{a-b}{a}=2$ [Option (1)].

4. To show that the circles always pass through fixed points, express the circle with center $(a,-a)$ and radius $r$:

   $$(x-a)^2+(y+a)^2 = r^2.$$

   But from (1) (with $b=-a$):

   $$(a+3)^2+a^2 = r^2+4 \quad \Rightarrow \quad r^2 = (a+3)^2+a^2-4.$$

   Thus, the circle is:

   $$(x-a)^2+(y+a)^2 = (a+3)^2+a^2-4.$$

   For a point $(x,y)$ to be common to all such circles (i.e. fixed), the relation must hold for all $a$. Writing the expanded form:

   $$x^2 - 2ax + a^2 + y^2 + 2ay + a^2 = x^2+y^2+2a^2+2a(-x+y).$$

   Since the right side is $(a+3)^2+a^2-4 = 2a^2+6a+5$, equate the coefficients of like powers of $a$:

   • Coefficient of $a^2$: $2 = 2$ (satisfied)
   • Coefficient of $a$: $2(-x+y)=6$ ⟹ $y-x=3$
   • Constant term: $x^2+y^2=5$

   Thus, the fixed points satisfy:

   $$y=x+3 \quad \text{and} \quad x^2+(x+3)^2=5.$$

   Solving:

   $$2x^2+6x+9=5 \quad \Rightarrow \quad 2x^2+6x+4=0 \quad \Rightarrow \quad x^2+3x+2=0,$$ $$(x+1)(x+2)=0 \quad \Rightarrow \quad x=-1 \text{ or } -2.$$

   Then, $y=-1+3=2$ or $y=-2+3=1$. Thus, the fixed points are $(-1,2)$ and $(-2,1)$ [Option (4)].

5. Option (2) $\frac{a-b}{a}=3$ is not true since we found $\frac{a-b}{a}=2$.

Explanation (Minimal):

• Given circles: $(x+3)^2+y^2=4$ and $x^2+(y-3)^2=4$.
• Using orthogonality, equate $(a+3)^2+b^2 = a^2+(b-3)^2$ giving $a+b=0$ (i.e. center lies on $x+y=0$).
• Then, $\frac{a-b}{a} = \frac{2a}{a} = 2$.
• Eliminating parameter $a$ shows fixed points $(-1,2)$ and $(-2,1)$ lie on every such circle.