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Circle with y = x - 2 as normal, abscissa twice ordinate, touching the x-axis - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
For a circle, the line y=x−2y=x-2 is a normal, the abscissa of the centre is twice its ordinate, and the circle touches the xx-axis. The equation of the circle is
Ax2+y2−8x−4y+16=0x^2+y^2-8x-4y+16=0correct
Bx2+y2−8x−4y+4=0x^2+y^2-8x-4y+4=0
Cx2+y2−4x−2y+2y+4=0x^2+y^2-4x-2y+2y+4=0
Dx2+y2−4x−2y+16=0x^2+y^2-4x-2y+16=0
Solution
Step 1: A normal to a circle passes through its centre, so the centre lies on y=x−2y = x-2. Step 2: Let the centre be (2t, t)(2t,\,t), the abscissa being twice the ordinate.
t=2t−2⟹t=2t = 2t-2 \quad\Longrightarrow\quad t = 2
 so the centre is (4,2)(4,2). Step 3: The circle touches the xx-axis, so the radius equals the distance of the centre from that axis, r=∣2∣=2r = |2| = 2. Step 4:
(x−4)2+(y−2)2=4(x-4)^2+(y-2)^2 = 4
x2−8x+16+y2−4y+4=4x^2-8x+16+y^2-4y+4 = 4
x2+y2−8x−4y+16=0x^2+y^2-8x-4y+16 = 0
 Step 5: Check. With g=−4, f=−2, c=16g=-4,\ f=-2,\ c=16 the xx-axis tangency test is g2=cg^2=c, and 16=1616=16. The radius is g2+f2−c=16+4−16=2\sqrt{g^2+f^2-c}=\sqrt{16+4-16}=2. Answer: (1) x2+y2−8x−4y+16=0x^2+y^2-8x-4y+16=0
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