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Question 106 of 126

Q.The general equation of a circle with centre (−3,−4)(-3, -4) and radius 3 units is :

(a) x2+y2−6x+8y−16=0x^2+y^2-6x+8y-16=0
(b) x2+y2−6x−8y+16=0x^2+y^2-6x-8y+16=0
(c) x2+y2+6x−8y+16=0x^2+y^2+6x-8y+16=0
(d) x2+y2+6x+8y+16=0x^2+y^2+6x+8y+16=0
Puducherry TnboardTamil Nadu HSC (DGE) Board 2022MCQ· 1mImportance★★★★★
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Expanding the standard-form circle (x+3)2+(y+4)2=9(x+3)^2+(y+4)^2=9 gives the general equation x2+y2+6x+8y+16=0x^2+y^2+6x+8y+16=0.

  1. The standard form of a circle with centre (h,k)(h,k) and radius rr is (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2.
  2. Here (h,k)=(−3,−4)(h,k)=(-3,-4) and r=3r=3, so (x−(−3))2+(y−(−4))2=32(x-(-3))^2+(y-(-4))^2=3^2, i.e. (x+3)2+(y+4)2=9(x+3)^2+(y+4)^2=9. …

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