Question

find the radius of the circle whose centre
is (3,2) and
passes through (√5,, 6). (pls answer with proper explanation )​

Answer 1

Given:

Centre point of circle = (3,2)

A point on circle = ($\sqrt{5}$,6)

To Find:

The radius of the circle

Concept:

The radius is the distance between the centre and a point on circle.

Distance Formula

Distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ units

Explanation:

Let the centre point be O(3,2) and A($\sqrt{5}$,6).

Radius = OA

In the question,

• $x_1$ = 3
• $y_1$ = 2
• $x_2$ = $\sqrt{5}$
• $y_2$ = 6

Substituting the values in Distance Formula, we get:

OA = $\sqrt{(\sqrt{5}-3)^2+(6-2)^2}$ units

We know that, $(a-b)^2=a^2+b^2-2ab$

OA = $\sqrt{5+9-6\sqrt{5}+16}$ units

OA = $\sqrt{30-6\sqrt{5}}$ units

We know that,

6$\sqrt{5}$ = 13.416407865

OA = $\sqrt{30-13.416407865}$ units

OA = $\sqrt{16.583592135}$ units

OA = 4.0722956836 units

Other Formulas:

1) Slope of Line

• Slope of a non-vertical line passing through points A(x_1,y_1) and B(x_1,y_2) is:

$\:\:\:\:\:\:\:\:\:\:\:\:m=\dfrac{y_2-y_1}{x_2-x_1}$

• If a line makes an angle \thetaθ with the positive side of x-axis, then the slope of line is:

$\:\:\:\:\:\:\:\:\:\:\:\:m = \tan\theta$

2) Equation of a Line

• Equation of a line parallel to x-axis at a distance b is:

$\:\:\:\:\:\:\:\:\:\:\:\:$y = by=b (where b is constant)

• Equation of a line parallel to y-axis at a distance is:

$\:\:\:\:\:\:\:\:\:\:\:\:$x = ax=a (where a is constant)

• Equation of a line having a slope and making an intercept with y-axis is:

$\:\:\:\:\:\:\:\:\:\:\:\:$y = mx+c (where m is the slope and c is the y-intercept made by line)

• Equation of a line when the line is passing through one point and slope is given:

$\:\:\:\:\:\:\:\:\:\:\:\:(y-y_1)=m(x-x_1)$(where $x_1,y_1$ are co-ordinates of point through which line passes and m is the slope).

• Equation of a non-vertical line passing through two points is:

$\:\:\:\:\:\:\:\:\:\:\:\:(y-y_1)=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$ (where $(x_1,y_1)\:and\:(x_2,y_2)$ are co-ordinates of two points through which line passes).

Conditions for two lines to be:

• Parallel is that the slope of both lines should ve equal.

$\:\:\:\:\:\:\:\:\:\:\:\:$ Let the slope of first line and second line be $m_1\:$and $m_2$ respectively.

Therefore, the two lines are parallel if $m_1=m_2$

• Perpendicular is that the product of the slopes of the two lines should be equal to -1.

$\:\:\:\:\:\:\:\:\:\:\:\:$ Let the slope of first and second line be $m_1$ and $m_2$ respectively.

Therefore, the two lines are perpendicular if $m_1\times m_2=-1$.