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:T480,{"@context":"https://schema.org","@type":"Quiz","name":"Mht Cet 2026 Circle PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/mht-cet/questions/2026/mathematics/circle/jt-mr4onrg9","educationalLevel":"Entrance exam preparation","about":"Mht Cet / Mathematics / Circle","hasPart":{"@type":"Question","name":"Mht Cet 2026 Circle PYQ with Solution","text":"The equation of the circle concentric with circle x^2 + y^2 - 6x + 7 = 0 and which touches the line x + y + 3 = 0 is ....","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"B"},"comment":{"@type":"Comment","text":"The given circle is x^2+y^2-6x+7=0 Complete the square for x: x^2-6x+y^2+7=0 (x-3)^2-9+y^2+7=0 (x-3)^2+y^2=2 So, the centre of this circle is (3,0) A concentric circle has the same centre. So its equation will be of the form x^2+y^2-6x+c=0 Now this circle touches the line x+y+3=0 So, its radius is equal to the perpendicular distance from the centre (3,0) to this line. Distance of point (x_1,y_1) from line ax+by+c=0 is \\frac{|ax_1+by_1+c|}{\\sqrt{a^2+b^2}} Here, \\frac{|3+0+3|}{\\sqrt{1^2+1^2}} = \\..."}}}18:T445,

The given circle is

$$ x^2+y^2-6x+7=0 $$

Complete the square for $x$:

$$ x^2-6x+y^2+7=0 $$

$$ (x-3)^2-9+y^2+7=0 $$

$$ (x-3)^2+y^2=2 $$

So, the centre of this circle is

$$ (3,0) $$

A concentric circle has the same centre. So its equation will be of the form

$$ x^2+y^2-6x+c=0 $$

Now this circle touches the line

Circle

Continue this paper

Similar Circle PYQs

More practice from this question