Question

in the given circle, C (-1,-4) is tge centre and AT is tangent to the circle . If A (5,1) and T (k,-2), then find the value of k

Answer 1

k = 27/5  or 2 ± √15

Step-by-step explanation:

Center of Circle  C = (- 1 , - 4)

its not clear which point is touching circle/ (tangent point) A or T

Let say point touching circle is A ( 5 , 1)

then Slope of AT  & AC  = -1  ( as AT ⊥ CA)

T = ( k , -2)

Slope of AT  =  (-2 - 1)/(k - 5)  = -3/(k - 5)

Slop of AC = (-4 - 1)/(-1 - 5) = -5/-6  = 5/6

(-3/(k - 5) )  * (5/6) = -1

=> 5k -25  = 2

=> k = 27/5

if  point touching circle is T ( k , -2)

Slope of AT * Slope of TC = -1

Slope of AT  =  (-2 - 1)/(k - 5)  = -3/(k - 5)

Slop of TC = (-4 - (-2))/(-1 - k) = -2/-(1 + k)  = 2/(k + 1)

-3/(k - 5)   * 2/(k + 1) = -1

=> (k - 5)(k + 1) = 6

=> k² - 4k - 5 = 6

=> k² - 4k - 11  =0

=> k =(4 ± √16 + 44)/2  =  2 ± √15

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Center of Circle C = (- 1 , - 4)

its not clear which point is touching circle/ (tangent point) A or T

Let say point touching circle is A ( 5 , 1)

then Slope of AT & AC = -1 ( as AT ⊥ CA)

T = ( k , -2)

Slope of AT = (-2 - 1)/(k - 5) = -3/(k - 5)

Slop of AC = (-4 - 1)/(-1 - 5) = -5/-6 = 5/6

(-3/(k - 5) ) * (5/6) = -1

=> 5k -25 = 2

=> k = 27/5

if point touching circle is T ( k , -2)

Slope of AT * Slope of TC = -1

Slope of AT = (-2 - 1)/(k - 5) = -3/(k - 5)

Slop of TC = (-4 - (-2))/(-1 - k) = -2/-(1 + k) = 2/(k + 1)

-3/(k - 5) * 2/(k + 1) = -1

=> (k - 5)(k + 1) = 6

=> k² - 4k - 5 = 6

=> k² - 4k - 11 =0

=> k =(4 ± √16 + 44)/2 = 2 ± √15