Equation of circle orthogonal with $2$ given circles

Question

Find the minimum radius of a circle which is orthogonal with both the circles: $C_1: x^2+y^2-12x+35=0$ and $C_2: x^2+y^2+4x+3=0$.

I know about the condition for a circle to be orthogonal to a given circle. $$d^2 = r_1^2+r_2^2,\;\text{ whereas }\;d = \overline{O_1O_2}$$

I couldn't apply this in $3$ circles. Moreover, i am unaware of the concept of radical and coaxial system.

I'll be obliged if someone could help.

Answer 1

first given a circle $C$ of unit radius, you can draw a circle $C_{\perp}$ centred at any point $P$ outside $C$. all you need to do is draw the two tangents to $C$ from $P$ let the contact points be $T_1, T_2.$ the radius is the common value $PT_1 = PT_2$ with center $P.$

suppose you have two non intersecting circles $C_1, C_2$ of equal radius $1$ and you require a circle $C_{\perp}$ orthogonal to both. by the previous argument, the center of $C$ is equal distance, the radius of $C_{\perp}.$ that means the center $P$ of $C_{\perp}$ is on the perpendicular bisector of the centers of $C_1$ and $C_2.$ the smallest radius occurs when $P$ is the midpoint of the centers of $C_1$ and $C_2$

the midpoint of the two centers $(6,0), (-2,0)$ of $x²+y²-12x+35=0 $ and $ x²+y²+4x+3=0$ is $(2,0)$ so the smallest radius is $$\sqrt{4^2 - 1} = \sqrt{15} $$

added later:

the reason for the above formula is that if the circle centered at $(2,0)$ has radius $r$ that is orthogonal to the circle of radius $1$ centered at $(6,0.)$ the distance between the centers is $4.$ therefore $r^2 + 1 = 4^2$

Answer 2

Hint: Find the centers and the radii of $C_1$ and $C_2$. Then find where a center of a circle orthogonal to both of these circles could lie.

Answer 3

Given the questions you have asked earlier, it looks like you are picking them up from fitjee RSM. But either way, to solve the question quickly, you should know the formula:

$2*(g1)*(g2) + 2*(f1)*(f2) = c1 + c2$ (where equation of circle: $x^2 + y^2 + 2gx+ 2fy + c =0$)

for first circle, $2*(-6)*f+0=35+c$

for second circle, $2*(2)*f=3+c$

solving, we get $f=-2$ and $c=-11$

Radius of circle is $(f^2+g^2-c)^{1/2} = (15+g^2)^{1/2}$

Now for minimum radius of circle, $g=0$, so radius = $\sqrt{15}$

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