The equation of the circle which touches the circle {\S} \equiv x^2+y^2-10 x-4 y+19=0 at the point (2,3) internally and having radius equal to half of the radius of the circle {\S}=0 is

Correct Answer is : x2+y2−7x−5y+16=0

TG EAMCET 4 May 2025 Shift 2 Paper

Question : 49 of 160

Marks: +1, -0

The equation of the circle which touches the circle S≡x2+y2−10x−4y+19=0 at the point (2,3) internally and having radius equal to half of the radius of the circle S=0 is

‌x2+y2+7x+5y+64=0
‌x2+y2−7x−5y+16=0
‌x2+y2−14x−10y+16=0
‌x2+y2−5x−7y+16=0

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