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# Find the equation of a circle. (i) which touches both the axes at a distance of 6 units from the origin. (ii) which touches x-axis at a distance 5 from the origin and radius 6 units. (iii) which touches both the axes and passes through the point (2, 1). (iv) passing through the origin, radius 17 and ordinate of the centre is -15.

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Solution

## The circle touches the axes at (0, 6) and (6, 0) respectively Thus, the centre of circle will be (6,6) and radius =OA=√(6−0)2+(6−6)2=√36=6 (by distance formula) ∴ the equation of circle will be (x−6)2+(y−6)2=62 ⇒x2+y2−12x−12y+36=0 (i) The circle touches the x - axis at A = (5, 0) and has radius 6 unit Thus, centre = (5, b) By distance formula OA = 6 ⇒√(5−5)2+(b−0)2=6⇒b=6⇒centre=(5,6) So, the equation of required circle is (x−5)2+(y−6)2=62⇒x2+y2−10x−12y+25=0 (iii) The circle touches both thb axis at A =(a, 0) and B (0, a) so, the centre of circle will be (a, a) and radius= a so, the equation of circle is (x−a)2+(y−a)2a2 …(A) Now, (A) Passes through P(2, 1) ∴(2−a)2+(1−a)2=a2⇒4−4a+a2+1−2a+a2=a2⇒5−6a+a2=0⇒a=5 or 1 Thus the equation of circle will be x2+y2−10x−lOy+25=0orx2+y2−2x−2y+1=0 (iv) The circle passes through origin (0, 0) and has radius = 17 units Also, die ordinate of centre is -15 then assume abssisa is a ∴OC=17⇒√(a−0)2+(0−15)2=17 (By distance formula) ⇒√a2+225=17⇒a2+225=289⇒a2=64⇒a=±8∴centre=(±8,−15) Thus the equation of circle will be, (x±8)2+(y+15)2=172⇒x2+y2±16x+30y=0  Suggest Corrections  2      Similar questions  Related Videos   The Fundamental Theorem of Arithmetic
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