Question Statement

The three lines $2x-y+1=0$, $2x+y-3=0$, and $x-2y+4=0$ touch a circle. Find the centre of the circle and the equation of the circle.

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Background and Explanation

To solve this problem, we use the property that the perpendicular distance from the centre of a circle $(h,k)$ to any of its tangent lines is equal to the radius $r$. By equating the distances from the centre to the three given lines, we can solve for the coordinates of the centre and subsequently find the radius and the circle's equation.

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Solution

Let $C(h,k)$ be the centre of the circle. Let the lines be:

1. $L_1:2x-y+1=0$
2. $L_2:2x+y-3=0$
3. $L_3:x-2y+4=0$

These lines touch the circle at points $P,Q,$ and $R$ respectively.

Since the distance from the centre to the tangent lines is equal to the radius, we have: $|CP|=|CQ|=|CR|$

Finding the Centre $C(h,k)$

Step 1: Equate $|CP|$ and $|CQ|$ Using the perpendicular distance formula $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$:

$\frac{|2h-k+1|}{\sqrt{2^2+(-1{)}^2}}=\frac{|2h+k-3|}{\sqrt{2^2+1^2}}$

$\Rightarrow \frac{2h-k+1}{\sqrt{5}}=\frac{2h+k-3}{\sqrt{5}}$

By equating the numerators: $2h-k+1=2h+k-3$ $\Rightarrow -k+1=k-3$ $\Rightarrow 2k=4$ $\Rightarrow k=2$

Step 2: Equate $|CP|$ and $|CR|$ $\frac{|2h-k+1|}{\sqrt{2^2+(-1{)}^2}}=\frac{|h-2k+4|}{\sqrt{1^2+(-2{)}^2}}$

$\Rightarrow \frac{2h-k+1}{\sqrt{5}}=\frac{h-2k+4}{\sqrt{5}}$

$\Rightarrow 2h-k+1=h-2k+4$ $\Rightarrow h+k=3$

Step 3: Substitute the value of $k$ Substitute $k=2$ into the equation $h+k=3$: $h+2=3$ $\Rightarrow h=1$

Thus, the centre of the circle is $C(1,2)$.

Finding the Radius $r$

The radius $r$ is the distance from $C(1,2)$ to any of the lines (e.g., $L_1$): $r=|CP|=\frac{|2(1)-2+1|}{\sqrt{2^2+(-1{)}^2}}$ $r=\frac{1}{\sqrt{5}}$

Finding the Equation of the Circle

Using the standard form $(x-h{)}^2+(y-k{)}^2=r^2$: $(x-1{)}^2+(y-2{)}^2={\left(\frac{1}{\sqrt{5}}\right)}^2$

Expand the terms: $x^2-2x+1+y^2-4y+4=\frac{1}{5}$ $x^2+y^2-2x-4y+5=\frac{1}{5}$

Multiply the entire equation by 5 to clear the fraction: $5(x^2+y^2-2x-4y+5)=1$ $5x^2+5y^2-10x-20y+25=1$ $5x^2+5y^2-10x-20y+24=0$

The required equation of the circle is $5x^2+5y^2-10x-20y+24=0$.

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Key Formulas or Methods Used

• Perpendicular Distance Formula: The distance from point $(x_1,y_1)$ to line $ax+by+c=0$ is $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$.
• Circle Property: The distance from the centre to any tangent line is equal to the radius ($r$).
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the centre and $r$ is the radius.

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Summary of Steps

1. Define the centre of the circle as $C(h,k)$.
2. Set up distance equations from $C(h,k)$ to the three given lines.
3. Equate the distances to solve for $k$ and then $h$.
4. Calculate the radius $r$ by substituting the centre coordinates back into the distance formula.
5. Plug the centre $(1,2)$ and radius $1/\sqrt{5}$ into the standard circle equation.
6. Simplify the equation into general form.