Question

prove that the tangent at (3,-2) to the circle x^2+y^2 =13 touches the circle x^2 +y^2 +2x-10y-26=0 and find the point of contact . ​

Answer 1

Step-by-step explanation:

given,

p(3,-2)

x^2+y^2=13

s=x^2+y^2+2x-10y-26=0

x^2+y^2=13

2x+2yy^1=0

y^1=-x/y

slope of tangent at (3,-2)=-3/-2=3/2

equation of tangent

(y+2)=3/2(x-3)

2y+4=3x-9

3x-2y=13

y=3x-13/2

x^2+(3x-13/2)^2+2x-10(3x-13/2)-26=0

x=5,y=1

(x,y)=(5,1)

hope it helps you❤

Answer 2

Given:

Equation of circle 1 is $x^{2} +y^{2} =13$

Equation of circle 2 is $x^{2} +y^{2} +2x-10y-26=0$

To find:

The point at which the tangent to the circle 1 and 2 touches the second circle.

Explanation:

• The general equation of a circle is $\sqrt{(x-h)^{2} +(y-k)^{2} } =R$ where, $(h,k)$ is the center and $R$ is the radius of the circle.

• Now, if the center of the circle lies at the origin, the equation of circle becomes  $\sqrt{x^{2} +y^{2}} =R$ .

• Now, if the center of the circle lies at an arbitrary point $(h,k)$, the equation is $x^{2}+ y^{2}+2gx+2fy+c=0$  where, $h=-g$   $;$ $k=-f$ and the radius for this circle is given by $R=\sqrt{g^{2}+ f^{2}-c }$.

Solution:

Now, in the question, we have been given that a tangent, lets say $l$ touches the circle 1 with center $O$ at $P(3,-2)$ and from the equation of the circle that is $x^{2} +y^{2} =13$ it is clear that the radius of the circle is $\sqrt{13}$ and the center of the circle lies at the origin $O(0,0)$.

Hence, the line $l$ is perpendicular to the circle 1 at the point of contact P.

We can take the slope of both the perpendicular lines that is, the line $OP$ and the tangent $l$.

Therefore,

Slope of line $OP$ ,  $m_{1} =\frac{-2-0}{3-0}=\frac{-2}{3}$

We know, the product of slopes of two mutually perpendicular lines is $-1$.

Let slope of the tangent $l$ be $m_{2}$

$m_{1}.m_{2}=-1$

$m_{2} =\frac{3}{2}$

Now, the tangent $l$ passes through $(3,-2)$ and has a slope $\frac{3}{2}$. Hence, the equation of tangent is given by,

$(y-y_{1} )=m(x-x_{1} )$

$y-(-2)=\frac{3}{2}(x-3)$

$2(y+2)=3(x-3)$

$2y+4=3x-9$

$3x-2y=13$

Hence, the equation of the tangent $l$ is $3x-2y-13=0$.

Now,

We have been given that the tangent $l$ touches the circle 2 with center O' given by $x^{2} +y^{2} +2x-10y-26=0$.

Here, $g=1$     $;$  $f=-5$     $;$  $c=-26$

Hence,

The radius of the circle is given by,

$R=\sqrt{(1)^{2}+ (5)^{2}-(-26) }$  

$R=\sqrt{26+26}$

$R=2\sqrt{13}$ $units$

Let, the tangent touches the circle at  $P'(x_{1}, y_{1} )$  and the distance of this point from the center of the circle $O'(-1,5)$  is nothing else but the radius of the circle whose value is $2\sqrt{13}$ units.

Using distance of a point from a point formula, we get

$\frac{|3x_{1} -2y_{1} -13|}{\sqrt{(3)^{2}+ (-2)^{2} } } =2\sqrt{13}$

$3x_{1} -2y_{1} -13=$ ± $26$  

$3x_{1} -2y_{1} =39$                              $(A)$

$3x_{1} -2y_{1}= -13$                           $(B)$                

We know, the slope of the radius of the circle is $\frac{-2}{3}$ and the center of the circle is $(-1,5)$. Hence, the equation will be,

$(y_{1} -5)=\frac{-2}{3}(x_{1} -(-1))$

$3(y_{1}-5) =-2(x_{1} +1)$

$3y_{1} -15=-2x_{1} -2$

$2x_{1} +3y_{1} =13$                             $(ii)$

Now,

Solving $(A)$ and $(ii)$, we get

$(x_{1}, y_{1} )=(11,-3)$

Solving $(B)$ and $(ii)$, we get

$(x_{1}, y_{1} )=(-1,5)$

Since the center and the circumference of the circle cannot lie on a single point. Hence, the correct point will be $(11,-3)$

Final answer:

Hence, the tangent touches the circle at $(11,-3)$.