Question

The line y=x+1 intersects the circlex2+y2-4x-2y+10, at points A and B. If C is the centre of the circle, find the: Centre and radius of the circle; Co-ordinates of A and B. The end points of a diameter of a circle are (6, 2) and(4,-3). Find the equation of the circle.​

Answer 1

Answer:

$(x - 2)^{2} + (y - 1)^{2} = 17$

Step-by-step explanation:

The current equation of the circle is:

⇒ $x^{2} + y^{2} - 4x - 2y + 10 = 0$

In order to get it into the standard form;

⇒ $(x - a)^{2} + (y - b)^{2} = r^{2}$

We must complete the square;

⇒ $(x - 2)^{2} - 4 + (y - 1) - 1 + 10 = 0$

Now, collect like terms and rearrange;

⇒ $(x - 2)^{2} + (y - 1)^{2} = -5?$

We now know that the Centre is at the point (2, 1).

We can use the distance formula to find the radius;

⇒ $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

⇒ $d = \sqrt{(6 - 2)^{2} + (2 - 1)^{2}}$

⇒ $\sqrt{17}$

Therefore the radius squared is 17.

Now substitute into our equation:

⇒ $(x - 2)^{2} + (y - 1)^{2} = 17$