TS EAMCET · Maths · Ellipse

If a circle $(x - 1)^{2} + y^{2} = r^{2}$ touches the ellipse $x^{2} + 4 y^{2} = 16$ internally, then $r =$

A $\sqrt{\frac{11}{3}}$
B $\frac{11}{3}$
C $\sqrt{\frac{15}{2}}$
D $2$

Verified Solution

Answer & Solution

Correct Answer

(A) $\sqrt{\frac{11}{3}}$

Step-by-step Solution

Detailed explanation

Given circle is x-12+y2=r2....i Here, centre of circle is 1,0 &radius=r. Given ellipse is x2+4y2=16 i.e. x216+y24=1 Comparing with x2a2+y2b2=1 a=4, b=2 Also given that the circle touches the ellipse internally as shown in figure. Let any point on ellipse be…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If a circle x-12+y2=r2 touches the ellipse x2+4y2=16 internally, then r=

Step-by-step Solution

If a circle $(x - 1)^{2} + y^{2} = r^{2}$ touches the ellipse $x^{2} + 4 y^{2} = 16$ internally, then $r =$