TS EAMCET · Maths · Circle

The number of common tangents that can be drawn to the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is

A $4$
B $0$
C $2$
D $1$

Verified Solution

Answer & Solution

Correct Answer

(A) $4$

Step-by-step Solution

Detailed explanation

For x2+y2=1 Centre C1≡0,0, Radius R1=1 For x2+y2-2x-6y+6=0 Centre C2≡1,3, Radius R2=12+32-6=2 C1C2=1-02+3-02=10 C1C2>3 ⇒C1C2>R1+R2 Circles are non-intersecting. So, there are 4 possible common tangents.

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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The number of common tangents that can be drawn to the circles x2+y2=1 and x2+y2-2x-6y+6=0 is

Step-by-step Solution

The number of common tangents that can be drawn to the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is