If the line 3x - 4y - k = 0 (k 0) touches t | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 4x - 8y - 5 = 0$. The centre is $(2, 4)$ and the radius is $r = \sqrt{2^2 + 4^2 - (-5)} = \sqrt{4 + 16

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 - 4x - 8y - 5 = 0$. The centre is $(2, 4)$ and the radius is $r = \sqrt{2^2 + 4^2 - (-5)} = \sqrt{4 + 16 + 5} = 5$.Since the line $3x - 4y - k = 0$ is a tangent,the perpendicular distance from the centre $(2, 4)$ to the line is equal to the radius $5$.$\frac{|3(2) - 4(4) - k|}{\sqrt{3^2 + (-4)^2}} = 5$$\frac{|6 - 16 - k|}{5} = 5$$|-10 - k| = 25$Since $k > 0$,we have $|-(10 + k)| = 25$,so $10 + k = 25$,which gives $k = 15$.The equation of the tangent is $3x - 4y - 15 = 0$.The normal at $(a, b)$ passes through the centre $(2, 4)$ and is perpendicular to the tangent. The slope of the tangent is $3/4$,so the slope of the normal is $-4/3$.The equation of the normal is $y - 4 = -\frac{4}{3}(x - 2)$,which simplifies to $4x + 3y = 20$.Solving the system of equations:$3x - 4y = 15$ $(i)$$4x + 3y = 20$ (ii)Multiplying $(i)$ by $3$ and (ii) by $4$: $9x - 12y = 45$ and $16x + 12y = 80$.Adding them gives $25x = 125$,so $x = a = 5$.Substituting $x = 5$ into (ii): $4(5) + 3y = 20$,so $3y = 0$,$y = b = 0$.Thus,$k + a + b = 15 + 5 + 0 = 20$.

If the line $3x - 4y - k = 0 (k > 0)$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$ at $(a, b)$,then $k + a + b$ is equal to:

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