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Question: Find the tangents of gradient \(\dfrac{4}{3}\), to the circle \({x^2} + {y^2} - 10x + 24y + 69 = 0\)...

Find the tangents of gradient 43\dfrac{4}{3}, to the circle x2+y2−10x+24y+69=0{x^2} + {y^2} - 10x + 24y + 69 = 0 and the coordinates of the points of contact.

Explanation

Solution

Hint: To solve this problem, let the point of contact be (h, k). We will put this point in the equation of circle and also apply conditions of two perpendicular lines m1m2=−1{m_1}{m_2} = - 1 for tangent and radius line to find the value of h and k and then find the equation of tangents by using the formula for equation of a line.

Complete step-by-step answer:
Now as we know that the standard equation of the circle is x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0. And the centre of this circle is (–g, –f) and its radius is g2+f2−c\sqrt {{g^2} + {f^2} - c} .
So, now the equation of the given circle is x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0.
So, comparing the given equation of the circle with the standard equation. We get the radius of the circle equal to 10 units and the centre as C (5, -12).
And as we know that the gradient of the tangent to the circle is the slope of the tangent.
Let the point of contact be P (h, k). Now, we will draw a figure according to the question. The equation of the circle is x2+y2−10x+24y+69=0{x^2} + {y^2} - 10x + 24y + 69 = 0. The centre has coordinates (5, -12) and the radius is 10 units. Slope of tangent = 43\dfrac{4}{3}

As we know that the slope of line joining two points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is calculated as m=y2−y1x2−x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
Now, slope of PC = −12 - k5 - h\dfrac{{ - 12{\text{ - k}}}}{{{\text{5 - h}}}}
Now, as we know that the tangent at point of contact is perpendicular to the radius of a circle.
And if two lines are perpendicular and their slopes are m1{m_1} and m2{m_2}. Then m1m2=−1{m_1}{m_2} = - 1
Therefore, (slope of tangent) x (slope of PC) = –1
Therefore, (−12 - k5 - h)43 = - 1\left( {\dfrac{{ - 12{\text{ - k}}}}{{{\text{5 - h}}}}} \right)\dfrac{4}{3}{\text{ = - 1}}
On simplifying the above expression, we get
3h + 4k = -33
h = −33 - 4k3{\text{h = }}\dfrac{{ - 33{\text{ - 4k}}}}{3} … (1)
Now, the point P lies on the circle. So, it also satisfies the equation of circlex2 + y2 - 10x + 24y + 69 = 0{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ - 10x + 24y + 69 = 0}}. Now, the equation of circle can be written as
(x - 5)2 + (y + 12)2 = 100{{\text{(x - 5)}}^2}{\text{ + (y + 12}}{{\text{)}}^2}{\text{ = 100}}
Therefore, (h - 5)2 + (k + 12)2 = 100{{\text{(h - 5)}}^2}{\text{ + (k + 12}}{{\text{)}}^2}{\text{ = 100}}
Now, putting value of h from equation (1), we get
(−33 - 4k3 - 5)2 + (k + 12)2 = 100{{\text{(}}\dfrac{{ - 33{\text{ - 4k}}}}{3}{\text{ - 5)}}^2}{\text{ + (k + 12}}{{\text{)}}^2}{\text{ = 100}}
(−48 - 4k3)2 + (k + 12)2 = 100{{\text{(}}\dfrac{{ - 48{\text{ - 4k}}}}{3}{\text{)}}^2}{\text{ + (k + 12}}{{\text{)}}^2}{\text{ = 100}}
169(k +  12)2 + (k + 12)2 = 100\dfrac{{16}}{9}{({\text{k + }}{\text{ 12)}}^2}{\text{ + (k + 12}}{{\text{)}}^2}{\text{ = 100}}
25(k + 12)2 = 90025{({\text{k + 12)}}^2}{\text{ = 900}}
(k + 12)2 = 36{({\text{k + 12)}}^2}{\text{ = 36}}
Taking square root both sides, we get
k + 12 = ±6{\text{k + 12 = }} \pm 6
Now, solving for both negative and positive sign, we get
k = -18, k = -6
Therefore, putting value of k in equation (1), we get
When k = -18
h = 13
When k = -6
h = -3
so, points of contact are (-3, -6) and (13, -18)
Now, let the equation of tangent is y = mx + c, where m = 43\dfrac{4}{3}.
Now, we will put the values of x and y and get the value of c.
Putting x = -3 and y = -6, we get
-6 = -3(43\dfrac{4}{3}) + c
c = -2
Therefore, the equation of tangent is y = 43\dfrac{4}{3}x – 2 i.e. 4x – 3y – 6 = 0
Again, putting x = 13 and y = -18, we get
-18 = 13(43\dfrac{4}{3}) + c
c = −1063\dfrac{{ - 106}}{3}
Therefore, the equation of tangent is y = 43\dfrac{4}{3}(x) + −1063\dfrac{{ - 106}}{3} i.e. 4x – 3y – 106 = 0

Note: When we come up with such types of questions, we will first find the centre of the circle and its radius by comparing the given equation of the circle with the standard equation of the circle. After that we put the assumed coordinates of the point of contact in the equation of the circle to form the equation in terms of h and k. After that we will form another equation by using the slope property of two perpendicular lines and after solving these two equations, we will get the two required value of h and k. now we can easily find the equation of tangent by using one point slope formula which states that that the equation of line passing through (x1,y1)\left( {{x_1},{y_1}} \right) and has slope m will be (y−y1)=m(x−x1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right). This will be the easiest and efficient way to find the solution of the problem.