Question

find the equation of circle which touches x axis at (3,0) and passes through (1,4)

Answer 1

Since circle is touching X axis at (3,0)
so centre must be (3,k).

And it is passing through( 1,4)

by using distance formula-
$\sqrt{ {(k - 4)}^{2} + {(3 - 1)}^{2} } = \sqrt{ {(k)}^{2} + 0} \\ sq. \: both \: sides - \\ {(k - 4)}^{2} + 4 = {k}^{2} \\ \\ 16 - 8k + 4 = 0 \\ k = \frac{5}{2} \\ \\ centre \: = (3, \: \frac{5}{2} )$
radius is 3

$so \: now \: eq. \: will \: be - \\ {(x - 3)}^{2} + {(y - \frac{5}{2}) }^{2} = {3}^{2}$

Answer 2

Answer:

$\text{Equation of circle : }(x-3)^2+(y-\frac{5}{2})^2=3^2$

Step-by-step explanation:

The circle is touching x - axis at (3,0)

So, its center will be of the form : (3,k)

The circle passes through (1,4). Now to find the value of k : distance between (3,0) and (3,k) is same as distance between (3,k) and (1,4) because (3,k) is the center of the circle.

So, using distance formula,

$\sqrt{(k-4)^2+(3-1)^2}=\sqrt{(k-0)^2+(3-3)^2}\\\text{On squaring both the sides}\\(k-4)^2+4=k^2\\16-8\cdot k+4=0\\\implies k=\frac{5}{2}\\\\\implies centre=(3,\frac{5}{2})\\\\radius=3\\\\\text{So, Equation of circle : }(x-3)^2+(y-\frac{5}{2})^2=3^2$