Question

The line $4x-3y+2 = 0$ intersects the circle $x^2+y^2-2x+6y+c=0$ at two points A, B and AB=8. If (1,k) is a point on the given circle and $k>0$, then $k =$

• A. 8
• B. 4
• C. 2
• D. 1

Hint:

The relationship between the radius ($r$) of a circle, the length of a chord ($L$), and the perpendicular distance from the center to the chord ($d$) is given by $r^2 = d^2 + (L/2)^2$. This is a direct application of the Pythagorean theorem and is very useful.

Answer 1

The Correct Option is C

Step 1: Find the properties of the circle.
The equation of the circle is $x^2+y^2-2x+6y+c=0$.
The centre of the circle is $C(-g, -f) = C(1, -3)$.
The radius is $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+3^2-c} = \sqrt{1+9-c} = \sqrt{10-c}$.
Step 2: Use the length of the chord to find the radius.
The line is $L: 4x-3y+2=0$. The length of the chord AB is given as 8.
Let $d$ be the perpendicular distance from the centre C(1,-3) to the line L.
$d = \frac{|4(1)-3(-3)+2|}{\sqrt{4^2+(-3)^2}} = \frac{|4+9+2|}{\sqrt{16+9}} = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3$.
Let M be the midpoint of the chord AB. We have a right-angled triangle CMA where $CA=r$, $CM=d$, and $AM = AB/2$.
$AM = 8/2 = 4$.
Using Pythagoras' theorem: $r^2 = d^2 + (AM)^2$.
$r^2 = 3^2 + 4^2 = 9+16 = 25$. So, the radius is $r=5$.
Step 3: Find the value of c.
We have $r^2 = 10-c$. So, $25 = 10-c \implies c = 10-25 = -15$.
The equation of the circle is $x^2+y^2-2x+6y-15=0$.
Step 4: Find the value of k.
The point (1,k) lies on the circle, so it must satisfy the circle's equation.
$1^2 + k^2 - 2(1) + 6(k) - 15 = 0$.
$1 + k^2 - 2 + 6k - 15 = 0$.
$k^2 + 6k - 16 = 0$.
Factoring the quadratic equation: $(k+8)(k-2)=0$.
The possible values for k are $k=-8$ or $k=2$.
The problem states that $k>0$, so we must choose $k=2$.