Question

In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.

92.$x^2+y^2=10;$at$(1,3)$.

Short answer

The equation of tangent line is$x+3y-10=0$.

Step-by-step solution

Step 1. Given information

The equation of tangent line must ne in the form of the $y=mx+b$.

Substitute $(x,y)=(1,3)$

$$\begin{gathered} 3=m\left(1\right)+b \\ 3=m+b \\ 3-m=b \\ b=3-m \end{gathered}$$

Now Substitute $b=3-m$in $y=mx+b$.

$y=mx+(3-m)$.

Step 2. Substitute y=mx+3-m in given equation x2+y2=10.

We get

$$\begin{gathered} x^2+{[mx+(3-m\left)\right]}^2=10 \\ {x}^2+m^2{x}^2+2mx(3-m)+{(3-m)}^2=10 \\ {x}^2(1+m^2)+6mx-2m^{2}x+9-6m+m^2=10 \\ {x}^2(1+m^2)+x(6m-2m^2)+(m^2-6m-1)=10 \end{gathered}$$

Using the quadratic formula,

$x=\frac{-(6m-2m^2)\pm \sqrt{{(6m-2m^2)}^2-4(1+m^2)(m^2-6m-1)}}{2(1+m^2)}$

To obtain a unique solution, the two roots must be equal which means

$$\begin{gathered} {(6m-2m^2)}^2-4(1+m^2)(m^2-6m-1)=0 \\ (36m^2-24m^3+4m^4)-4(m^2-6m-1)-4(m^4-6m^3-m^2)=0 \\ 36m^2+24m+4=0 \\ 9m^2+6m+1=0 \\ {(3m+1)}^2=0m=-\frac{1}{3} \end{gathered}$$

Step 3. Substitute m=-13 in y=mx+b.

We get

$$\begin{gathered} y=-\frac{1}{3}x+b \\ y=-\frac{1}{3}x+(3+\frac{1}{3}) \\ 3y=-x+10 \\ 3y+x-10=0 \\ x+3y-10=0 \end{gathered}$$

So the required equation of a tangent line of$x^2+y^2=10$at$(1,3)$is$x+3y-10=0$.$(1,3)$