EllipsemediumFree

Eccentricity of Centre Locus of a Circle | JEE

JEE Maths question with a full step-by-step solution.

Question
A circle S=0S=0 touches the circle x2+y2−4x+6y−23=0x^2+y^2-4x+6y-23=0 internally and the circle x2+y2−4x+8y+19=0x^2+y^2-4x+8y+19=0 externally. The locus of the centre of S=0S=0 is a conic whose eccentricity is kk. Then [1k]\left[\dfrac1k\right] (greatest integer function) is:
A77correct
B22
C00
D33
Solution
Step 1: First circle: centre C1=(2,−3)C_1=(2,-3), radius r1=4+9+23=6r_1=\sqrt{4+9+23}=6. Second circle: centre C2=(2,−4)C_2=(2,-4), radius r2=4+16−19=1r_2=\sqrt{4+16-19}=1. Step 2: Let CC be the centre and rr the radius of S=0S=0. Internal tangency to circle 1: CC1=r1−rCC_1=r_1-r. External tangency to circle 2: CC2=r2+rCC_2=r_2+r. Adding,
CC1+CC2=r1+r2=7 (constant).CC_1+CC_2=r_1+r_2=7\ \text{(constant)}.
 So the locus is an ellipse with foci C1,C2C_1,C_2 and 2a=72a=7. Step 3: 2ae=C1C2=0+1=12ae=C_1C_2=\sqrt{0+1}=1, so
e=2ae2a=17.e=\frac{2ae}{2a}=\frac{1}{7}.
 Step 4: Here k=e=17k=e=\dfrac17, so 1k=7\dfrac1k=7 and [1k]=7\left[\dfrac1k\right]=7. Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Ellipse · easy
The equation (x²)/(10-a)+(y²)/(4-a)=1 represents an ellipse if:
Ellipse · easy
The eccentricity of the ellipse (x-3)²+(y-4)²=(y²)/(9) is:
Ellipse · medium
An ellipse having foci at (3,3) and (-4,4) and passing through the origin has eccentricit…
Ellipse · medium
The foci of an ellipse are (-2,4) and (2,1) . The point (1,(23)/(6) ) is an extremity of…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath