Question

Equation of DCT is y+4= -(x - 22)
24
= 24y +96 = 7 x - 154 = 7x – 24 y - 250 = 0)
**13. Find the transverse common tangents of t
circles
x² + y2 - 4x - 10y + 28 = () and
x2 + y2 + 4x – 6y+4= 0
(Mar-2017, 2014, 2013, 2008, 20
ol: Given equation of the circles are​

Answer 1

Answer:

1

:x

2

+y

2

+6x+4y+4=0,C

1

=(−3,−2),r

1

=3

S

2

:x

2

+y

2

–2x=0,C

2

=(1,0),r

2

=1

The external center of similitude cuts the line joining the centers in the ratio r

1

:r

2

externally

P=

r

1

–r

2

r

1

C

1

–r

2

C

2

=(

2

3–(−3)

,

2

0−(−2)

)=(3,1)

Any tangent to S

2

is given by

y=m(x–1)+

m

2

+1

–(1)

(3,1) passes through the above line

⟹1=m(3–1)+

m

2

+1

4m

2

+1–4m=m

2

+1

3m

2

=4m

m=0,

3

4

Substituting in (1)

The common tangents are

y=1,4x–3y−9

(II)

S

1

:(x−2)

2

+(y–5)

2

=1,C

1

=(2,5),r

1

=1

S

2

:(x+2)

2

+(y–3)

2

=9,C

2

=(−2,3),r

2

=3

Any tangent to S

1

is given by

(y–5)=m(x–2)±1

m

2

+1

y=mx+(5±

m

2

+1

–2m)–(1)

Any tangent to S

2

is given by

(y–3)=m(x+2)±3

m

2

+1

y=mx+(3+2m±3

m

2

+1

)–(2)

(1) And (2) represent the same line

⟹5±

m

2

+1

–2m=3+2m±3

m

2

+1

4m–2=−2

m

2

+1

16m

2

+4–8m=2m

2

+2

8m

2

–8m+2=0

⟹m=

2

1

(or)

4m–2=4

m

2

+1

⟹m=

4

−3

The internal central of similitude is given by

Q=

r

1

+r

2

r

1

C

2

+r

2

C

1

=(

4

4

,

4

18

)=(1,

2

9

)

Using (1) and verifying slopes substituting Q we get the common tangents as

x=1,3x+4y–21=0

Answer 2

Answer:

Answer:

1

:x

2

+y

2

+6x+4y+4=0,C

1

=(−3,−2),r

1

=3

S

2

:x

2

+y

2

–2x=0,C

2

=(1,0),r

2

=1

The external center of similitude cuts the line joining the centers in the ratio r

1

:r

2

externally

P=

r

1

–r

2

r

1

C

1

–r

2

C

2

=(

2

3–(−3)

,

2

0−(−2)

)=(3,1)

Any tangent to S

2

is given by

y=m(x–1)+

m

2

+1

–(1)

(3,1) passes through the above line

⟹1=m(3–1)+

m

2

+1

4m

2

+1–4m=m

2

+1

3m

2

=4m

m=0,

3

4

Substituting in (1)

The common tangents are

y=1,4x–3y−9

(II)

S

1

:(x−2)

2

+(y–5)

2

=1,C

1

=(2,5),r

1

=1

S

2

:(x+2)

2

+(y–3)

2

=9,C

2

=(−2,3),r

2

=3

Any tangent to S

1

is given by

(y–5)=m(x–2)±1

m

2

+1

y=mx+(5±

m

2

+1

–2m)–(1)

Any tangent to S

2

is given by

(y–3)=m(x+2)±3

m

2

+1

y=mx+(3+2m±3

m

2

+1

)–(2)

(1) And (2) represent the same line

⟹5±

m

2

+1

–2m=3+2m±3

m

2

+1

4m–2=−2

m

2

+1

16m

2

+4–8m=2m

2

+2

8m

2

–8m+2=0

⟹m=

2

1

(or)

4m–2=4

m

2

+1

⟹m=

4

−3

The internal central of similitude is given by

Q=

r

1

+r

2

r

1

C

2

+r

2

C

1

=(

4

4

,

4

18

)=(1,

2

9

)

Using (1) and verifying slopes substituting Q we get the common tangents as

x=1,3x+4y–21=0