Question

7) Find the equation of circle (a) passing through
the origin and having intercepts 4 and -5 on
the coordinate axes​

Answer 1

Find the equation of the circle which touches the X axis at the point (3,0) and cuts off an intercept 8 units from the positive part of the y axis

Answer by Edwin McCravy(17323) (Show Source): You can put this solution on YOUR website!

Find the equation of the circle which touches the X axis at the point (3,0) and cuts off an intercept 8 units from the positive part of the y axis

The standard equation of any circle is

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

Since it is tangent to the x-axis at (3,0),

the center is directly above (3,0), so its

center also has x-coordinate 3. So h=3

%28x-3%29%5E2%2B%28y-k%29%5E2=r%5E2

It passes through (x,y) = (3,0), so

%283-3%29%5E2%2B%280-k%29%5E2=r%5E2

0%5E2%2Bk%5E2=r%5E2

k%5E2=r%5E2

k and r are both positive numbers.

k=r and

%28x-3%29%5E2%2B%28y-r%29%5E2=r%5E2

It has a y-intercept at (0,8)

%280-3%29%5E2%2B%288-r%29%5E2=r%5E2

%28-3%29%5E2%2B64-16r%2Br%5E2=r%5E2

9%2B64-16r%2Br%5E2=r%5E2

73-16r=0

-16r=-73

r=%28-73%29%2F%28-16%29

r=73%2F16

And since x=r=73%2F16,

%28x-3%29%5E2%2B%28y-73%2F16%29%5E2=%2873%2F16%29%5E2

AMP Parsing Error of 'drawing(4800/13,400,-3,9, -3,10,circle(3, 73/16,.1),

locate(3,73/16,(matrix(1,3,3,',',73/16))),

graph(4800/13,400,-3,9, -3,10), circle(3,73/16,73/16) )': Invalid expression '',',73/16))),graph(4800/13,400,-3,9,-3,10),circle(3,73/16,73/16))': syntax error at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 194. .

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Answer 2

The required equation of the circle is $x^2+y^2-4x+5y= 0$

Step-by-step explanation:

Given:

Here,

Let OA is chord of the given circle.

And we know that perpendicular drawn from center of circle to the chord bisects the chord.

so, L is the midpoint of OA.  

And OA = 4

then, OL= $\frac{OA}{2}$ =$\frac{4}{2}$ = 2

And since OL = 2 which means x−coordinate of point C is 2.

Similarly perpendicular drawn from C to chord OB bisects the chord OB.

suppose perpendicular drawn from center C to chord OB intersects at point P.

OB = -5

Then, OP = $\frac{OB}{2}$ = $\frac{-5}{2}$

So, CL = OP = $\frac{-5}{2}$

And since CL= $\frac{-5}{2}$ which means y−coordinate of point C is $\frac{-5}{2}$.

Therefore, center is $C (2,\frac{-5}{2} )$

Using Pythagoras theorem, we have

$OC^2= OL^2+CL^2$

               = $2^2+ (\frac{-5}{2} )^2$    [∵ CL = OP = $\frac{-5}{2}$]

               = $4+\frac{25}{4}$

               = $\frac{16+25}{4}$

               = $\frac{41}{4}$

               = $r^2$

Also, coordinates of the centre of the circle are $(2,\frac{-5}{2} )$. Thus, the required equation of the circle is,

$(x-2)^2+(y-(\frac{-5}{2} ))^2 = r^2$

$x^2-4x+4+(y+\frac{5}{2} ))^2 = \frac{41}{4}$

$x^2-4x+4+y^2+2\times\frac{5}{2}\times y+\frac{25}{4} = \frac{41}{4}$

$x^2-4x+4+y^2+5y +\frac{25}{4}-\frac{41}{4}= 0$

$x^2-4x+4+y^2+5y +\frac{25-41}{4}= 0$

$x^2-4x+4+y^2+5y +\frac{-16}{4}= 0$

$x^2+y^2-4x+5y= 0$

Hence the required equation of the circle is $x^2+y^2-4x+5y= 0$.