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Question 43 of 47

Q.The equation of the circle with centre on the xx-axis and passing through the origin is :

(a) x2+y2=a2x^2+y^2=a^2
(b) x2−2ax+y2=0x^2-2ax+y^2=0
(c) x2−2ay+y2=0x^2-2ay+y^2=0
(d) y2−2ay+x2=0y^2-2ay+x^2=0
Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2025MCQ· 1mImportance★★★★★
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Centre (a,0)(a,0), radius aa; (x−a)2+y2=a2(x-a)^2+y^2=a^2 simplifies to x2−2ax+y2=0x^2-2ax+y^2=0.

Step 1 — Set up. The centre lies on the xx-axis, so it is of the form (a,0)(a,0). The circle passes through the origin, so the radius equals the distance from (a,0)(a,0) to (0,0)(0,0), i.e. r=ar=a.

Step 2 — Write the equation.

(x−a)2+(y−0)2=a2.(x-a)^2+(y-0)^2=a^2.

Step 3 — Expand.

x2−2ax+a2+y2=a2.x^2-2ax+a^2+y^2=a^2. …

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