Question

Suppose \(d_1\) and \(d_2\) are respectively the lengths of intercepts of the circles \(x^2+y^2=4\) and \(x^2+y^2-10x-14y+65=0\) on the line \(2x-2y-3=0\). Then which of the following is true?

• A. \(d_1=2d_2\)
• B. \(d_2=2d_1\)
• C. \(d_1=3d_2\)
• D. \(d_1=d_2\)

Hint:

The length of chord intercepted by a line on a circle is \[ 2\sqrt{r^2-p^2} \] where \(r\) is the radius of the circle and \(p\) is the perpendicular distance from the centre to the line.

Answer 1

The Correct Option is D

Step 1: Recall the formula for length of intercept made by a circle on a line.
If a circle has radius \(r\), and the perpendicular distance of its centre from the line is \(p\), then the length of chord intercepted by the line is \[ d=2\sqrt{r^2-p^2} \]

Step 2: Find \(d_1\) for the circle \(x^2+y^2=4\).
The circle is \[ x^2+y^2=4 \] Its centre is \[ (0,0) \] and radius is \[ r_1=2 \] The given line is \[ 2x-2y-3=0 \] Distance of \((0,0)\) from the line is \[ p_1=\frac{|2(0)-2(0)-3|}{\sqrt{2^2+(-2)^2}} \] \[ p_1=\frac{3}{\sqrt8} \] Therefore, \[ d_1=2\sqrt{2^2-\left(\frac{3}{\sqrt8}\right)^2} \] \[ d_1=2\sqrt{4-\frac{9}{8}} \] \[ d_1=2\sqrt{\frac{23}{8}} \]

Step 3: Find \(d_2\) for the circle \(x^2+y^2-10x-14y+65=0\).
The equation can be written as \[ x^2-10x+y^2-14y+65=0 \] Completing squares, \[ (x-5)^2-25+(y-7)^2-49+65=0 \] \[ (x-5)^2+(y-7)^2=9 \] Thus, the centre is \[ (5,7) \] and radius is \[ r_2=3 \] Distance of \((5,7)\) from the line \(2x-2y-3=0\) is \[ p_2=\frac{|2(5)-2(7)-3|}{\sqrt{2^2+(-2)^2}} \] \[ p_2=\frac{|10-14-3|}{\sqrt8} \] \[ p_2=\frac{7}{\sqrt8} \] Therefore, \[ d_2=2\sqrt{3^2-\left(\frac{7}{\sqrt8}\right)^2} \] \[ d_2=2\sqrt{9-\frac{49}{8}} \] \[ d_2=2\sqrt{\frac{23}{8}} \]

Step 4: Compare \(d_1\) and \(d_2\).
We have \[ d_1=2\sqrt{\frac{23}{8}} \] and \[ d_2=2\sqrt{\frac{23}{8}} \] Hence, \[ d_1=d_2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{d_1=d_2} \]