Grade Select GradeIIT JEE Entrance Exam

lines 5x+12y-10=0 & 5x-12y-40=0 touch a circle of diameter 6 find equatoin of another circle which is concentric with the given circle and c uts intercepts of lenght 8 on these lines

LK Shrinidhi

11 Years agoGrade Select Grade

1 Answer

Arun

6 Years ago

If C be the centre of the circle C

then it lies on the bisectors of the given lines which are

5x+12y−10

=±

5x−12y−40

These give x=5 and y=−5/4. Since centre lies in 1st quadrant therefore y=−5/4 is ruled out. Let the centre be (5,k) then its perpendicular distance from each of lines will be 3,

i.e. p=r.

∴

25+12kl−10

=3 and

25−12k−40

∴k=2 or k=−

=−

The value −9/2 of k is ruled out as centre C lies in 1st quadrant ∴C is (5,2).

Now the circle C

cuts off intercepts 8 from these lines and if its radius be r, then

=25 or r=5

∴ Circle C

is (x−5)

+(y−2)

or x

−10x−4y+4=0.

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lines 5x+12y-10=0 & 5x-12y-40=0 touch a circle of diameter 6 find equatoin of another circle which is concentric with the given circle and c uts intercepts of lenght 8 on these lines

LK Shrinidhi

1 Answer

Arun

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