Question
The radii of the circles $x^{2}+y^{2}=1, x^{2}+y^{2}-2 x-6 y-6=0$ and $x^{2}+y^{2}-4 x-12 y-9=0$ are in(a) G.P(b) A.P(c) H.P(d) AGP
Step 1
Step 1: The general equation of a circle is given by $x^{2}+y^{2}+2gx+2fy+c=0$, where the center of the circle is $(-g,-f)$ and the radius is $\sqrt{g^{2}+f^{2}-c}$. Show more…
Show all steps
Your feedback will help us improve your experience
Saurabh Chandra and 77 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The two circles $x^{2}+y^{2}=a x$ and $x^{2}+y^{2}=c^{2}(c>0)$ touch each other if (A) $|a|=c$ (B) $a=2 c$ (C) $|a|=2 c$ (D) $2|a|=c$
The circles $x^{2}+y^{2}-x-y=0$ and $x^{2}+y^{2}-x+y=0$ intersect at an angle (a) $\frac{\pi}{2}$ (b) $\frac{\pi}{3}$ (c) $\frac{\pi}{6}$ (d) $\frac{\pi}{4}$
The locus of the centres of the circles which touch the two circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4 a x$ externally is (A) $12 x^{2}-4 y^{2}-24 a x+9 a^{2}=0$ (B) $12 x^{2}+4 y^{2}-24 a x+9 a^{2}=0$ (C) $12 x^{2}-4 y^{2}+24 a x+9 a^{2}=0$ (D) none of these
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD