Question

What is the equation of the chord centered at (1, 2) in the circle $x^2+y^2-4x-6y-10=0$

• A. $x-y-3=0$
• B. $x-y+3=0$
• C. $x+y-3=0$
• D. $x+y+3=0$

Answer 1

The correct option is C

$x+y-3=0$

This can be solved in 2 methods.

Method - 1

Given circle is

, $x^2+y^2-4x-6y-10=0$

$(x-2{)}^2+(y-3{)}^2=10+9+4$

= 23

This is shown in the figure. P is the given point and O is the centre of the circle.

Also AB is the required chord.

Slope of $PO=\frac{3-2}{2-1}=1$

$\therefore$ slope of AB $=-1,$ because $AB\perp OP.$

$\therefore$ Required equation of AB is,

$y-2=(-1)(x-1)=0$

i.e., $y-2=-x+1$

i.e., $x+y-3=0$

Method 2

In this we use the formula to find the equation of the chord.

$\to T_1=S_1$

Where $T_1$ is obtained by replacing
$x^2\text{by}x{x}_1,y^2\text{by}y{y}_1,x\text{by}(\frac{x+x_1}{2}),y\text{by}(\frac{y+y_1}{2})$
$S_1$ is obtained by replacing $x$ by $x_1$ and $y$ by $y_1.$

$\therefore T_1=S_1$

$x+2y-2(x+1)-3(y+2)-10=1+4-4-12-10$

$-x-y-8=-11$

i.e.,$x+y-3=0$