Question
If the circles $x^{2}+y^{2}=\lambda^{2}$ and $x^{2}+y^{2}-6 x-8 y+9=0$ touch externally, then $\lambda$ is(a) $-2$(b) 3(c) 2(d) 1
Step 1
The first circle has center at $(0,0)$ and radius $\lambda$. The second circle can be rewritten in the form $(x-3)^2+(y-4)^2=4^2$ which has center at $(3,4)$ and radius $4$. Show more…
Show all steps
Your feedback will help us improve your experience
Saurabh Chandra and 88 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 circle $x^{2}+y^{2}+2 \lambda x=0, \lambda \in R$ touches the parabola $y^{2}=4 x$ externally only if (a) $\lambda>0$ (b) $\lambda<0$ (c) $\lambda>1$ (d) $\lambda=1$
If a circle passes through the points of intersection of the coordinate axes with the lines $\lambda x-y+1=0$ and $x-2 y+3=0$, then the value of $\lambda$ is (A) 2 (B) 1 (C) $-1$ (D) $-2$
If $\lambda x^{2}-10 x y+12 y^{2}+5 x-16 y-3=0$ represents a pair of straight lines, then the equation of one of them is (a) $x-2 y+3=0$ (b) $x-2 y-1=0$ (c) $2 x-6 y+3=0$ (d) $2 x-6 y-1=0$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD