Question

0. Equation of the circle which is such that the lengths of the tangents to it from the points (1, 0), (0, 2) and
(3, 2) are 1, root 7 and root 2 respectively ​

Answer 1

Answer:

Let centre be (x , y ) and radius r

Given

$(x - 1) {}^{2} + {(y - 0)}^{2} + {r}^{2} = {1}^{2} \\ (x - 0) {}^{2} + {(y - 2)}^{2} + {r}^{2} = 7 \\ {(x - 3)}^{2} + {(y - 2)}^{2} + {r}^{2} = 2$

Evaluate for x , y and r

( If you are fail to solve let me know I will help you)

Answer 2

Equation of the circle is x2+y2-(13/6)x-(7/12)y+13/3=0.

Step-by-step explanation:

Given:

Equation of the circle $x^2+y^2+2gx+2fy+c=0$

Length of the tangents are $1, \sqrt{7}, \sqrt{2}$

So point (1, 0) and length 1

⇒ $\sqrt{(1)^2+(0)^2+2(g)(1)+2(f)(0)+c} =1$

⇒$\sqrt{2g+c+1}=1$

⇒$2g+c+1=\sqrt{1}$

⇒2g+c+1=1

⇒c=-2g                   [1]

Point (0, 2) and length $\sqrt{7}$,

$\sqrt{(0)^2+(2)^2+2(g)(0)+2(f)(2)+c} =\sqrt{7}$

$\sqrt{4+4f+c} =\sqrt{7}$

4+4f+c=7

4f+c=7-4

4f+c=3                    [2]

Point (3, 2) and length $\sqrt{2}$,

$\sqrt{(3)^2+(2)^2+2(g)(3)+2(f)(2)+c}=\sqrt{2}$

$\sqrt{9+4+6g+4f+c}=\sqrt{2}$

13+6g+4f+c=2

6g+4f+c=2-13

6g+4f+c=-11            [3]

Subtracting equation 2 from 3

6g+4f+c-(4f+c)=-11-3

6g+4f+c-4f-c=-13

6g=-13

$g=\frac{-13}{6}$

Putting value of g in equation 1

c=-2g

c=$-2\frac{-13}{6}$

c=$\frac{13}{3}$

Putting value of g and c in equation 3

6g+4f+c=-11

$6(\frac{-13}{6} )+4f+\frac{13}{3} =-11$

$-13+4f+\frac{13}{3} =-11$

$4f=-11+13-\frac{13}{3}$

$4f=2-\frac{13}{3}$  

$4f=\frac{6-13}{3}$  

$f=\frac{6-13}{3\times 4}$  

$f=\frac{-7}{12}$  

Equation of the circle is,

$x^2+y^2+2gx+2fy+c=0$

$x^2+y^2-\frac{13}{6}x-\frac{7}{12}y+\frac{13}{3}=0$