AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The figure below represents the two circles with the common tangent.
The slope of the common tangent,
m=−‌

4

3

The slope of the line perpendicular to tangent is,
m′=tan‌θ=‌

3

4

Therefore, sin‌θ=‌

3

5

,cos‌θ=‌

4

5

Now,
‌

x−1

4∕5

=‌

y−1

3∕5

=±5
Now,
x=(±5×‌

4

5

+1),y=(±5×‌

3

5

+2)
x=(5,−3),y=(5,−1)
The coordinates of C1(5,5) and C2(−3,−1)
The equations of the required circles is,
(x−5)2+(y−5)2=52
x2+y2−10x−10y+25=0
And
(x+3)2+(y+1)2=52
x2+y2+6x+2y−15=0