If the equation of the tangent to the circle x-3-5 cos- y-1-5 sin- at the point 0-3 is p x-q y-r-0 and p-0 then p-q-

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Question

If the equation of the tangent to the circle x=3+5 $c o s θ$ , $y$ = $- 1 + 5 sin θ$ at the point (0,3) is $p x + q y + r$ =0 and $p > 0$ then p + q =

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Solution

The correct option is B
-1

x = 3 + 5 cos $θ$
y = - 1 + 5 sin $θ$
Equation of circle is
$(x - 3)^{2} + (y + 1)^{2} = 25 x^{2} + y^{2} - 6 x + 2 y - 15 = 0$
Equation of tangent at (0,3) is
$3 y - 3 x + y + 3 - 15 = 0 4 y - 3 x - 12 = 0 3 x - 4 y + 12 = 0 S o, p + q = - 1$

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If the equation of the tangent to the circle x=3+5cos θ, y=−1+5sinθ at the point (0,3) is px+qy+r =0 and p>0 then p + q =

The correct option is B -1 x = 3 + 5 cos θ y = - 1 + 5 sin θ Equation of circle is (x−3)2+(y+1)2=25x2+y2−6x+2y−15=0 Equation of tangent at (0,3) is 3y−3x+y+3−15=04y−3x−12=03x−4y+12=0So,p+q=−1

If the equation of the tangent to the circle x=3+5 $c o s θ$ , $y$ = $- 1 + 5 sin θ$ at the point (0,3) is $p x + q y + r$ =0 and $p > 0$ then p + q =

The correct option is B
-1

x = 3 + 5 cos $θ$
y = - 1 + 5 sin $θ$
Equation of circle is
$(x - 3)^{2} + (y + 1)^{2} = 25 x^{2} + y^{2} - 6 x + 2 y - 15 = 0$
Equation of tangent at (0,3) is
$3 y - 3 x + y + 3 - 15 = 0 4 y - 3 x - 12 = 0 3 x - 4 y + 12 = 0 S o, p + q = - 1$