16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4a7,{"@context":"https://schema.org","@type":"Quiz","name":"Ap Eapcet 2024 Circle PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/ap-eapcet/questions/2024/mathematics/circle/m88c6f6m","educationalLevel":"Entrance exam preparation","about":"Ap Eapcet / Mathematics / Circle","hasPart":{"@type":"Question","name":"Ap Eapcet 2024 Circle PYQ with Solution","text":"If \\theta is the angle between the tangents drawn from the point (2,3) to the circle x^2+y^2-6 x+4 y+12=0 then \\theta=","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"D"},"comment":{"@type":"Comment","text":"We have, Equation, of circle is x^2+y^2-6 x+4 y+12=0 Centre of circle =(\\lambda-2) radius =\\sqrt{9+4-12}=1 \\begin{aligned} O P & =\\sqrt{(3-2)^2+(-2-3)^2} \\\\ & =\\sqrt{1+25}=\\sqrt{26} \\\\ A P & =\\sqrt{O P^2-A O^2} \\\\ & =\\sqrt{26-1}=5 \\end{aligned} \\ln \\triangle M O P, \\tan \\frac{\\theta}{2}=\\frac{A O}{A P}=\\frac{1}{5} \\tan \\theta=\\frac{2 \\tan \\frac{\\theta}{2}}{1-\\tan ^2 \\frac{\\theta}{2}}=\\frac{2 \\times \\frac{1}{5}}{1-\\left(\\frac{1}{5}\\right)^2}=\\frac{\\frac{2}{5}}{\\frac{24}{25}}=\\frac{5}{12} Hence,..."}}}18:T44c,

We have,

Equation, of circle is

$$ x^2+y^2-6 x+4 y+12=0 $$

Centre of circle $=(\lambda-2)$

radius $=\sqrt{9+4-12}=1$

Circle

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