Pls answer 84 question
Q.84. If equation of one tangent drawn from (0, 0) to the circle with centre (2, 4) is 4x + 3y = 0, then equation of the other tangent from (0, 0) is
(1) 4x - 3y = 0
(2) x = 0
(3) y = 0
(4) x + 4y = 0



Dear Student,



Given : equation of one tangent from (0, 0) to the circle with centre (2, 4) is 4x+3y=0.To Calculate  : equation of the other tangent from (0,0).solution : radius of the circle will be length of perpendicular on line 4x+3y=0 from centre (2, 4).As We know,length of perpendicular  from a point (x1, y1) on the line ax+by+c=0 is = ax1+by1+ca2+b2Radius of the circle =  4(2)+3(4)+042+32=8+1216+9=205=4 unitsSo, equation of circle will be (x-2)2+(y-4)2=42x2+22-2×x×2+y2+42-2×y×4=16x2+y2-4x-8y+20-16=0S : x2+y2-4x-8y+4=0g=coefficient of x2=-42=-2f=coefficient of y2=-82=-4equation of pair of tangents from  a point P(x1, y1) is SS1=T2(x2+y2+2gx+2fy+c)(x12+y12+2gx1+2fy1+c)={(xx1+yy1+g(x+x1)+f(y+y1)+c}2Here S : x2+y2-4x-8y+4=0 and P(x1, y1)=(0, 0)(x2+y2-4x-8y+4)( 02+02-4×0-8×0+4)=(x×0+y×0+(-2)(x+0)+(-4)(y+0)+4)24(x2+y2-4x-8y+4)=(-2x-4y+4)24x2+4y2-16x-32y+16=4x2+16y2+16+2{(-2x)(-4y)+(-2x)4+(-4y)4}4x2+4y2-16x-32y+16=4x2+16y2+16+16xy-16x-32y4x2+4y2-16x-32y+16-4x2-16y2-16-16xy+16x+32y=0-12y2-16xy=0-4y(3y+4x)=0(4x+3y)y=0since one of the tangent is 4x+3y=0so other tangent will be y=0 or x axis.

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