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
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 .
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
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...
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
Similar questions
French,
8 months ago
Math,
8 months ago
Political Science,
8 months ago
Science,
1 year ago
Math,
1 year ago