Math, asked by Samraan, 1 year ago

What are all the common tangents of the circle x^2 + y^2 = 9 and x^2 + y^2 - 16x +2y +49 = 0

Answers

Answered by abhi178
7
first find centre of circles and radius of circles .

for equation :- x^2 + y^2 =9
centre (0,0) and radius 3 unit

for equation x^2 +y^2 -16x +2y +49 =0
centre ( 8, -1) and radius √(64 +1 -49) =4

now ,
distance between centre = √(64 +1) =√65

sum of radius = 3 + 4 = 7

here we see that ,
sum of radius < distance between centre

hence, four common tangents are possible . e.g two direct common tangents and two transverse common tangents .
Answered by kvnmurty
6
The questions should be given where the numbers should be easy to calculate.

x² + y² = 9    => C1 = O(0,0),  r1 = 3
x²+y²-16x +2y+49=0 => (x-8)²+(y+1)²=16  => C2= (8,-1),   r2 = 4

C1 C2 = √65 > r1 + r2.   Hence There are 4 tangents in total.
Equation of C1 C2:   x + 8 y = 0

   The points P and Q of intersection of direct and transverse common tangents lie on the line C1 C2.  By drawing a simple diagram, We can see that P lies to the left of C1 and Q lies to the right of C1. So:
     P = (-8a , a), a>0.     Q = (8b, -b) , b > 0

We use the ratios of distances (similar triangles formed by tangents).

C1 P = C1C2 * r1 / |r2 - r1| = √65 * 3/(4-3) = 3√65
    =>   √[(8a)²+a²] = 3√65    => a = 3
    =>   P = (-24, 3)

Let  P1 (x1, y1) be points of contact of direct tangents from P to circle C1.
    => PP1 ⊥ P1C1
    => (y1 -3)/(x1+24) * y1/x1 = -1   => x1²+y1² +24x1-3y1 = 0
    => 8 x1 - y1 + 3 = 0     or  y1 = 8 x1+3
Then  x1²+(8 x1+3)² = 9    => 65 x1² + 48 x1 = 0 
      => x1 = 0 or - 48/65
      => P1 = (0, 3 )  and  P2 = (-48/65, -189/65)

Direct Tangents:  x x1 + y y1 = 9
     =>  T1 :  y =3           ,  T2 = 16 x + 63 y = 195
=======
For Transverse tangents:   Point of intersection with C1 C2 = Q = (8b, -b).  Let P3(x3, y3) be the points of contact with Circle C1.
We use the ratio of distances and radii.

   C1Q = C1C2 * r1 /(r1+r2) = √65 * 3/7
   C1Q² = √65 b = √65 * 3/7    =>   b = 3/7       =>  Q = (24/7, -3/7)
 
QP3 ⊥ C1P3    =>  (y3+3/7)/(x3 - 24/7)  * y3/x3 = -1
      => y3² + x3² + 3 y3 /7 - 24 x3 /7 = 0
      => y3 - 8 x3 + 21 = 0          => y3 = 8 x3 - 21
      => x3² + (8 x3 - 21)² = 9
      => 65 x3² - 336 x3 + 432 = 0
      =>  x3 = [168 +- 12]/65 = 36/13  or 12/5
      =>  y3 = 15/13    or   -9/5
Points of contact:  P3 = (36/13, 15/13)  and  P4 = (12/5, -9/5)

Transverse tangents:   x x3 + y y3 = 9
     12 x + 5 y = 39       and      4 x - 3 y = 15

=== aaaahhhh! ...  Long sequence of steps and calculations...


kvnmurty: click on red heart thanks above pls
abhi178: excellent answer sir , if i asked this question , definitely mark this brainliest
Samraan: Wonderful answer sir, and the way of presentation was really helpful! Thanking you sincerely!
kvnmurty: thanks for selecting brainliest.
abhi178: thank samraan !!
Samraan: No worries brother! :)
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