Find the equation of direct common tangents of the circles x2 + y2 –
2x – 6y + 9 = 0, x2+ y2-4 = 0
Answers
Answer:
Step-by-step explanation:
The tangent equation to (x−a)2+(y−b)2=r2can be written as y−b=m(x−a)±r1+m2−−−−−−√
(If you are aware of this, proceed through the solution. If you are not, try to prove this, and if you don’t get it, comment below and I’ll be happy to help)
Circle 1: x2+y2=16−>
Tangent: y=mx±41+m2−−−−−−√
Circle 2: (x−2)2+(y−1)2=9−>
Tangent: y−1=m(x−2)±31+m2−−−−−−√=>y=mx+1−2m±31+m2−−−−−−√
Now since both the tangent equations should be same as they are common tangents, let us compare them. So,
±41+m2−−−−−−√=1−2m±31+m2−−−−−−√
=> ±1+m2−−−−−−√=1−2m
=> 1
+m2=1−4m+4m2
=> 3m2−4m=0=>m=0orm=4/3
Now substitute back m in both tangent equations, and pull out the intercept value which is common to both the equations. So you will the tangent equations as:
y=4
3y=4x−20
Hope this helped! If you have further doubts, feel free to drop a comment.
Edit:
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N.K Sharma
N.K Sharma, former Jo Prapt Hai Vo Prayapt Hai
Answered May 28, 2016 · Author has 906 answers and 1.1m answer views
Equations of given circles are
x^2+y^2=16...........................(1)
x^2+y^2-4x-2y-4=0................(2).
For cicle (1) center (0,0) radius=4
For circle (2)center(2,1) radius=3
Let y=mx+b be equation of common tangents to the given circles.
Thus c/√(1+m^2)=4................(4)
and (2m+c-1)/√(1+m^2)=3.....(5)
Eliminating c between (4) and (5) we get 1-2m=√(1+m^2)
1-4m+4m^2=1+m^2
m(3m-4)=0=>m=0,4/3 and c =0,20/3
Hence equations of the common tangents are
y=4,3x-4x=20
Hope you understand your point...
3.8k views · View 5 Upvoters · Answer requested by Ashwin Srikanth
Two methods
Method 1 : First find point of intersection of tangents by section formula. Use (T)^2=(S)*(S1) to get the equation of pair of tangents and then solve it to get two equations separately. This method can be used for both Transverse Common Tangent and Direct Common Tangent. But it is lengthy and hence I don't prefer this method
Method 2 : Assume the point of intersection of common tangents to be P(h, k). And then form equation of tangent as (y-k) =m(x-h). And then find out m. For this use the fact that distance of this line from the center circles is r1 and r2 respectively. You will ...
Shambhu Bhat
Shambhu Bhat, Retired professor in engineering ;Very fond of mathematics
Answered Mar 26, 2017 · Author has 2.3k answers and 1.1m answer views
x2+y2=9
is a circle with center C1=(0,0)
and radius=4
x2+y2–4x−2y−4=0
is a circle with center C2=(2,1)
and radius 3.
the 2 circles intersect and there are 2 common tangents.the point where the tangents meet ,P divides C1C2
externally in the ratio of 4:-3.
By section formula we get P(8,4)
Now we need to find equations of tangents from P.
Let y-4=m(x-8) be the equation of a tangent .
Then perpendicular distance of the line from C1
= radius 4
|0–4−m(0–8)|m2+1−−−−−−√=4
simplifying we get 3m2–4m=0
m=0 or m=43
Plugging these values the equations of tangents are y=4 and 4x-3y-20=0
Answer:
Step-by-step explanation:
find point of intersection of direct common tangents
point divides centres in the ratio of their radius
radius1=1
radius2 = 2
centre 1 = 2,3
centre 2 = 0,0
this shall be divided in the ratio of their radii
point of intersection = 4/3 , 3
y-3 = m(x-4/3)
3y-9 = m(3x-4)
3mx-3y+9-4m = 0
apply tangency condition
9-4m / root 2 = 2
(9-4m)^2 = 8
16m^2+81-72m = 8
16m^2-72m+73 = 0
this can be done by another way also
slope of line inclined theta with the other line = m+-tantheta / 1-+mtantheta
tantheta = r1-r2 / c1c2