Question

Find the value of " k" so that the straight line 4x+3y+k=0 may touch the circle X2+y2-4x+10y+4=0.

Answer 1

Given,

The equation of a circle:

$\rm \: {x}^{2} + {y}^{2} - 4x + 10y + 4 = 0 \\ \\ \rm \implies\: {x}^{2} + {y}^{2} + 2( - 2)x + 2 \times 5x + 4 = 0 \: \: \: - - - (1)$

We know that ,

Standard, equation of circle is :

$\rm \: a {x}^{2} + a {y}^{2} + 2gx + 2fy + c = 0$

its center is = (-g,-f)

$\sf \: and \: \: radius \: \: r = \sqrt{ {g}^{2} + {f}^{2} - c }$

Now, by comparing eqn(1) with standard equation, we will get,

g = -2

f = 5

c = 4

So,

center = (2,-5)

$\sf \: radius \: \: r = \sqrt{ { {( - 2)}^{ 2} } + {5}^{2} - 4} \\ \\ = \sqrt{4 + 25 - 4} \\ \\ = \sqrt{25} \\ \\ = 5 \: unit$

Again given that,

4x+3y+k=0 line touches the circle.

So, we can write that,

distance between center and line = radius of circle.

According to the question,

$\rm \: \frac{ |4 \times 2 + 3 \times ( - 5) + k| }{ \sqrt{ {4}^{2} + {3}^{2} } } = 5 \\ \\ \rm \implies\: \frac{ |8- 15 + k| }{ \sqrt{25} } = 5 \\ \\ \rm \implies\: \frac{ |k - 7| }{5} = 5 \\ \\ \rm \implies\: |k - 7| =5 \times 5 = 25 \\ \\ \rm \implies\:k - 7 = \pm \: 25 \\ \\ \sf \: taking \: + ve \\ \\ \rm \: k - 7 = 25 \\ \rm \implies\:k = 25 + 7 = 32 \\ \\ \sf \: taking \: - ve \\ \\ k - 7 = - 25 \\ \rm \implies\:k = - 25 + 7 = - 18$

So,

$\rm \: k = 32 \: \: or \: - 18$

Answer 2

Answer:

The values of k are 32, -18.

Step-by-step explanation:

General equation of circle:

    The general equation of circle is as

         x²+y²+2gx+2fy+c = 0 with

          center = (-g,-f) and radius = $\sqrt{g^{2}+f^{2}-c }$

Distance between a point and a line:

  Let the line equation be ax+by+c = 0 and the point = (x₁,y₁)

     then the distance between the point and the line is $\frac{|ax_{1}+by_{1}+c |}{\sqrt{a^{2}+b^{2} } }$.

    Given circle equation is

                x²+y²-4x+10y+4  = 0

       x²+y²+2(-2)x+2(5)y+4  = 0  -------(i)

    compare equation(i) with the general equation of circle we get

                    g = -2, f = 2, c = 4

          center = (-g,-f) = (2,-5)

           radius = $\sqrt{2^{2}+5^{2}-4 }$  

                      = √(4+25-4

                      = √25

          radius  =  5

• Given straight line is 4x+3y+k = 0 and the given condition is straight line touches the circle.

• If a line touches the circle then the radius of circle is equal to the distance from center of circle to the line equation.

• Distance from center (2,-5) of circle to the line is given by

           (|4(2)+3(-5)+k|)/√(4²+3²) = (|8-15+k|)/√(16+9)

                                                   = (|-7+k|)/(√25)

                                                   = |k-7|/5

             As, Radius = distance from center tot he circle

                            5 = |k-7|/5

                          25 = |k-7|

                        |k-7| = 25

                         k-7 = ±25

                         k-7 = +25   or   k-7 = -25

                           k = 25+7   or     k = -25+7

                           k = 32       or      k = -18

  Hence , the values of k are 32, -18.

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