Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the line  touches the circle  <img alt="x^2+y^2-4 x-8 y-5=

If the line 3 x-4 y-\lambda=0 touches the circle 

x^2+y^2-4 x-8 y-5=0 at (a,b), then \lambda+a+b is equal to 

 

Option: 1

20


Option: 2

22


Option: 3

-30


Option: 4

-28


Answers (1)

best_answer

Since the given line touches the given circle, the length of the perpendicular from the centre (2,4) of the circle from the line 3 x-4 y-\lambda=0 is equal to the radius \sqrt{(4+16+5)}=5 of the circle.

\Rightarrow \frac{(3 \times 2-4 \times 4-\lambda)}{\sqrt{(9+16)}}= \pm 5 \Rightarrow \lambda=15 \text { or }-35

Now, equation of the tangent at (a,b) to the given circle is 

\begin{aligned} & x a+y b-2(x+a)-4(y+b)-5=0 \\ \\\Rightarrow & (a-2) x+(b-4) y-(2 a+4 b+5)=0 \end{aligned}

If it represents the given line 3 x-4 y-\lambda=0, then 

\frac{a-2}{3}=\frac{b-4}{-4}=\frac{2 a+4 b+5}{\lambda}=l \text { (say) }

Then a=3 l+2, b=4-4 l \text { and } 2 a+4 b+5=\lambda l\: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)

\begin{aligned} & \Rightarrow \quad 2(3 l+2)+4(4-4 l)+5=15 l(\text { if } \lambda=15) \\ \\& \Rightarrow \quad l=1 \Rightarrow a=5, b=0 \text { and } \lambda+a+b=20 \end{aligned}

Again, if \lambda=-35                                                                              (from i)

\begin{aligned} & 25-10 l=-35 l \Rightarrow l=-1 \Rightarrow a=-1, b=8 \\ \\& \text { and } \lambda+a+b=-35-1+8=-28 \\ & \end{aligned}

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question