JEE Class main Answered

find number of common tangent of circle x²+y²=9 and (x-8)²+(y-10)²=25 and also find equation of transverse common tangent

Asked by ishantm672 | 02 Jan, 2023, 17:17: PM

Expert Answer

Let Circle-1 be ( x² + y² ) = 9 = 3²

Centre of circle C₁ (0, 0 ) and radius r₁ = 3

Let Circle-2 be ( x-8)² +(y-10)² = 25 = 5²

Centre of circle C₂ (8, 10 ) and radius r₂ = 5

Distance d between centres ,

begin mathsize 14px style d space equals space square root of left parenthesis space 8 minus 0 right parenthesis squared plus left parenthesis 10 minus 0 right parenthesis squared end root space equals space 12.81 end style

Distance d between centers of circles greater than sum of radius (3+5) , circles do not intersect.

Hence number of common tangents are 4 .

-------------------------------------------------------------------------------------

Transverse tangents AB and CD are shown in figure,

Let AB and CD intersect at P which is on the line joining the centres of circles.

C₁P : C₂P = r₁ : r₂ = 3 : 5

Coordinate of point of intersection P is

begin mathsize 14px style open square brackets fraction numerator 3 cross times 8 plus 5 cross times 0 over denominator 3 plus 5 end fraction space comma space fraction numerator 3 space cross times 10 space plus space 5 space cross times 0 over denominator 3 plus 5 end fraction close square brackets space equals space left parenthesis space 3 comma space 15 over 4 right parenthesis end style

Let the coordinate of A be ( x, y)

Equation of line AP :-

begin mathsize 14px style y space minus space 15 over 4 space equals space m space left parenthesis space x space minus space 3 right parenthesis end style

.......................................(1)

where m is slope of line AP. Equation of line AP is [ 4m x - 4y +(15-12m) ] = 0

Perpendicular distance r₁ from C₁ to the line AP is given as

begin mathsize 14px style r subscript 1 space equals space fraction numerator open vertical bar 15 minus 12 m close vertical bar over denominator square root of 16 m squared plus 16 end root end fraction space equals space 3 end style

(15-12m)² = 9 ( 16 m² + 16 )....................................(2)

(5-4m)² = (16m² + 16 )

from above expression, we get m = 9/40

By substituting m in eqn.(1) , we get line AP as , 9x -40y+123 = 0

In quadratic eqn.(2) , other value of m is

begin mathsize 14px style infinity end style

or m = 1/0

If we substitute m = 1/0 in eqn.(1) , we get other line CP as ( x-3 ) = 0

Hence equations of transverse common tangents are 9x -40y+123 = 0 and ( x-3 ) = 0

Answered by Thiyagarajan K | 03 Jan, 2023, 10:32: AM

Please Login or Sign up to continue!