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Mathematics Question on Conic sections

If the straight line 3x+4y=k3x + 4y = k touches the circle x2+y2=16xx^2 + y^2 = 16x, then the value of kk is

A

16,64

B

-16,-64

C

-16,64

D

16,-64

Answer

-16,64

Explanation

Solution

Given, x2+y2=16xx^{2} + y^{2} = 16x
⇒x2−16x+y2=0\Rightarrow x^{2} -16x + y^{2} = 0
⇒x2−2(x)8+(8)2−(8)2+y2=0\Rightarrow x ^{2}- 2\left(x\right) 8 + \left(8\right)^{2} - \left(8\right)^{2} + y^{2} = 0
⇒(x−8)2+y2=64...(i)\Rightarrow \left(x-8\right)^{2}+y^{2} =64 \quad...\left(i\right)
∴\therefore Centre = (8, 0) and radius = 8
Since, the straight line 3x+4y=k3x + 4y = k touches the circle x2+y2=16x^{2} + y^{2} = 16, therefore the length of perpendicular from the centre (8, 0) to the straight line 3x+4y−k=03x + 4y - k = 0 is equal to the radius of the circle.
i.e., 8=∣3(8)+4(0)−k9+16∣8 = \left|\frac{3\left(8\right)+4\left(0\right)-k}{\sqrt{9+16}}\right|
⇒8=∣24−k25∣\Rightarrow 8 = \left|\frac{24-k}{\sqrt{25}}\right|
⇒8=∣24−k∣5\Rightarrow 8 = \frac{\left|24 - k\right|}{5}
⇒∣24−k∣=40\Rightarrow \left|24-k\right| = 40
⇒24−k=±40\Rightarrow 24-k = \pm 40
Taking positive sign, we get
24−k=4024 -k =40
⇒−k=−40−24=16\Rightarrow - k = - 40 - 24 = 16
⇒k=−16\Rightarrow k = -16
Taking negative sign, we get
24−k=−4024-k=-40
⇒−k=−40−24=−64\Rightarrow -k =-40-24=-64
⇒k=64\Rightarrow k =64
∴k=−16, 64\therefore k =-16,\,64