equation of circle tangent to line with radius

Question

Find the equation of a circle tangent to line $3x + y - 2 = 0$ at $(-1,5)$ and with radius $\sqrt{10}$. I've no idea on how to do this.

Answer 1

Now as per the question the circle touches the line only at one point so it is a tangent and there can be two circles touching the given line and passing through (-1,5).

so if the given line is a tangent to the circle then the centre lies on a line perpendicular to the given line, passing through (-1,5) and at a distance of $\sqrt(10)$.

slope of line perpendicular to it is $$m1*m2=-1$$ where m1=-3 which implies m2 =1/3.

let the line be y=(1/3)x+c. Now this passes through (-1,5) substitute and get c value

The resulting equation which is perpendicular to the given line is $$3y=x+16---(1)$$

so to find the circle equation we need to find the centre of the circle which can be found with the equation shown which is nothing but the distance formula$$(x-(-1))^2+(y-5)^2=(\sqrt10)^2 ---(2)$$

since the circle has a radius of $\sqrt(10)$ units which implies the centre is at a distance of $\sqrt(10)$ units from it.

solving (1) and (2) we get $$x=-4,2$$ and $$y=4,6$$

therefore the equations of two circles are $$(x-(-4))^2+(y-4)^2=(\sqrt10)^2$$ and $$(x-2)^2+(y-6)^2=(\sqrt10)^2$$

hope this helped.

Answer 2

Let the equation of the circle be $$(x-h)^2+(y-k)^2=(\sqrt{10})^2$$

As the circle passes through $(-1,5);$ we have $$(-1-h)^2+(5-k)^2=(\sqrt{10})^2$$

So, the parametric form of $(h,k)$ can be $(\sqrt{10}\cos A-1,\sqrt{10}\sin A+5)\ \ \ \ (1)$

From the Article$\#148$ of Loney's Coordinate Geometry Book, the radius of a circle = the perpendicular distance of the centre from a tangent

So, we have $$\sqrt{10}=\frac{\left|3(\sqrt{10}\cos A-1)+(\sqrt{10}\sin A+5)-2\right|}{\sqrt{10}}$$

$$\iff|3\cos A+\sin A|=\sqrt{10}$$

Let $3=r\cos B,1=r\sin B\implies r=\sqrt{10}\implies|\cos(A-B)|=1\implies\cos(A-B)=\pm1$

Case $\#1:$ If $\cos(A-B)=1,A-B=2m\pi\implies \cos A=\cos B=\dfrac3{\sqrt{10}},\sin A=\sin B=\dfrac1{\sqrt{10}}$

Use $(1)$ to find $(h,k)$

Case $\#2:$ If $\cos(A-B)=-1,A-B=(2m+1)\pi\implies \cos A=\cdots=-\cos B,\sin A=-\sin B$

Use $(1)$ to find $(h,k)$

Answer 3

Let the equation be $$10=(x-a)^2+(y-b)^2$$

$$\implies10=(a+1)^2+(b-5)^2\ \ \ \ (1)$$

Again like equation of circle tangent to line with radius, $$\dfrac{3a+b-2}{\sqrt{3^2+1^2}}=\pm\sqrt{10}$$

$$3a+b-2=\pm10$$

Considering '+' sign, $$3a+b-2=10\iff b=12-3a\ \ \ \ (2)$$

Use $(1),(2)$ find $a$ and hence $b$

Similarly consider '-' sign