Question

a) Determine whether the given Line is a tangent to tre given circle.
*+ 2y +6 = 0, x + y2 - 6x – 4y + 8
b) Find the tangents from the origin to the circle x +y? - 10x - 6y + 25 = 0.
[Hint: Lei (a, b) be the point of contact of the tangent on the circle. The radius to (a, b) is
perpendicular to the tangent, this gives a quadratic in a and h. (a, b) is on the circle, this gives
Mother quadratic. Solve simultaneously.]​

Answer 1

Answer:

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Answer 2

Explanation 1

Given:

Equation of the circle = $x^{2}+ y^{2}-6x-4y+8$

A line $x+2y+6=0$

To find:

To find whether the line is a tangent to a circle or not.

Solution:

Step 1

We know, the equation of a circle is given by $x^{2}+ y^{2}+2gx+2fy+c=0$ where $R$ is the radius of circle and the radius is given by $R=\sqrt{g^{2}+ f^{2}-c }$

Now, equating the general equation with the given equation,

$g=-h$     and   $f=-k$  

We get $(h,k)=(3,2)$

$R=\sqrt{(3)^{2}+ (2)^{2}-8 }$

$R=\sqrt{5}$  ≅ $2.2units$

Step 2

Now, we have the given line $x+2y+6=0$  and the center of the circle $(3,2)$

Hence,

Distance of a point from the line $D=\frac{Ax_{1}+By_{1}+C }{\sqrt{A^{2}+ B^{2} } }$

Substituting the known values in the equation, we get

$D=\frac{3+2(2)+6}{\sqrt{(1)^{2}+ (2)^{2} } }$

$D=\frac{3+4+6}{\sqrt{1+4} }$

$D=$ $\frac{13}{\sqrt{5} }$  ≅ $5.8$

$5.8$ is a much greater value than the radius of the circle.

Hence, the given line passes through a much greater distance from the circumference of the circle and hence cannot touch the circle therefore, The line is not a tangent of the circle.

Final answer:

The line is not a tangent of the circle.

Explanation 2

Given:

Equation of circle $x^{2}+ y^{2}-10x-6y+25$

To find:

The tangents of the circle

Solution:

We already know the basic equation of a circle that is $x^{2}+ y^{2}+2gx+2fy+c$ . Equating this with the equation given to us, we get

$g=-5$    and  $f=-3$

or the center of the circle $(h,k)=(5,3)$

Radius $R=\sqrt{g^{2}+ f^{2}-c }$

$R=\sqrt{(5)^{2}+(3)^{2} -25 }$

$R=3units$

We can clearly see that the radius of the circle and the y coordinate of the center of circle are equal. Hence, we can understand that the circle passes through the x axis itself and the center of circle being at a distance of 5 units from the y axis.

Therefore,

The first tangent for the circle will be x axis itself that is $y=0$.

Let the equation of this tangent be $y=mx+c$ ; where $c$ is the intercept made by the line on the y axis.

The tangent here, does not make any intercept on the y axis

Therefore, $c=0$

Hence, the equation becomes $y=mx$

or $mx-y=0$    $(i)$

We can clearly see that, from a unique point where the tangent will touch the circle, the distance to the center will be nothing else but the radius of the circle which is $3units.$

We know,

Distance of a point from a line is $D=\frac{Ax_{1} +By_{1} +C}{\sqrt{A^{2}+ B^{2} } }$

Here, $(x_{1}, y_{1} )=(5,3)$ and $A=m$  $;$   $B=-1$   $;$   $C=0$

Substituting the given values, we get

$3=\frac{m(5)-1(3)+0}{\sqrt{m^{2}+1 } }$

$5m-3=3\sqrt{m^{2}+1 }$

Squaring both sides, we get

$25m^{2} -30m+9=9m^{2}+9$

$16m^{2}-30m=0$

$m(16m-30)=0$

$m=0$   $;$  $m=\frac{15}{8}$

Hence, the tangent becomes

$y=\frac{15}{8} x$

Final answer:

Hence, the two tangents for the circle are $y=0$ and $y=\frac{15}{8} x$.