Find the co-ordinate of the point on the circle `x^2+y^2-12x-4y +30=0`, which is farthest from the origin.

Options

A. (9, 3)
B. (8, 5)
C. (12, 4)
D. none of these

Answer

To find the coordinates of the point on the circle given by the equation \(x^2 + y^2 - 12x - 4y + 30 = 0\) that is farthest from the origin, we can follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
We start with the equation of the circle:
\[ x^2 + y^2 - 12x - 4y + 30 = 0 \]
To rewrite this in standard form, we complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 12x \quad \text{can be written as} \quad (x - 6)^2 - 36 \]
2. For \(y\):
\[ y^2 - 4y \quad \text{can be written as} \quad (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x - 6)^2 - 36 + (y - 2)^2 - 4 + 30 = 0 \]
Simplifying this, we get:
\[ (x - 6)^2 + (y - 2)^2 - 10 = 0 \quad \Rightarrow \quad (x - 6)^2 + (y - 2)^2 = 10 \]
This shows that the center of the circle is at \((6, 2)\) and the radius is \(\sqrt{10}\).

### Step 2: Determine the distance from the origin to the center of the circle
The distance \(d\) from the origin \((0, 0)\) to the center \((6, 2)\) can be calculated using the distance formula:
\[ d = \sqrt{(6 - 0)^2 + (2 - 0)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \]

### Step 3: Find the farthest point on the circle from the origin
To find the point on the circle that is farthest from the origin, we need to move from the center \((6, 2)\) outward in the direction of the origin. The farthest point will be along the line that connects the origin and the center of the circle.

1. The direction vector from the origin to the center is \((6, 2)\).
2. Normalize this vector to find the unit vector:
\[ \text{Unit vector} = \left(\frac{6}{\sqrt{40}}, \frac{2}{\sqrt{40}}\right) = \left(\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}}\right) \]

3. The farthest point from the origin will be at the center plus the radius in the direction of the unit vector:
\[ \text{Farthest point} = (6, 2) + \sqrt{10} \left(\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}}\right) = (6 + 3, 2 + 1) = (9, 3) \]

### Conclusion
The coordinates of the point on the circle that is farthest from the origin are \((9, 3)\).

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