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JEE MATHEMATICS MCQ: CIRCLES

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Equation of a circle with centre (4,3)(4,3) touching the circle x2+y2=1x^2+y^2=1 is
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Correct option: D
x2+y2−8x−6y+9=0x^2+y^2-8 x-6 y+9=0
Option D (x2+y2−8x−6y+9=0x^2+y^2-8 x-6 y+9=0) is the correct answer because it represents a circle with center (4,3) that touches the circle x2+y2=1x^2 + y^2 = 1 externally.

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The sum of the coefficients in the expansion of (α2x2−2αx+1)51,{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}}, as a polynomial in xx, vanishes. Position of the point (α,2α2)\left( {\alpha ,2{\alpha ^2}} \right) with respect to the circle x2+y2=4,{x^2} + {y^2} = 4, is One of the diameter of the circle x2+y2−2x+4y−4=0{x^2} + {y^2} - 2x + 4y - 4 = 0 is The locus of the middle point of the chords of the circle x2+y2=a2{x^2} + {y^2} = {a^2} such that the chords pass through a given point (x1,y1)\left( {{x_1},{y_1}} \right), is α,β\quad \alpha, \beta and γ\gamma are the parametric angles of three points P,QP, Q, and R respectively, on the circle x2+y2=1x^2+y^2=1, and AA is the point (−1,0)(-1,0). If the lengths of the chords AP,AQA P, A Q and ARA R are in GPG P, then cos⁡α/2,cos⁡β/2\cos \alpha / 2, \cos \beta / 2 and cos⁡γ/2\cos \gamma / 2 are in The radius of the circle passing through the point (6,2), two of whose diameters are x+y=6x+y=6 and x+2y=4x+2 y=4 is If a circle has two of its diameters along the lines x+y=5x+y=5 and x−y=1x-y=1 and has area 9π9 \pi, then the equation of the circle is
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