Prove that the tangent to the circle x2 + y2 = 5 at the point (1,-2 ) also touches the - Maths - Circles

Question

Prove that the tangent to the circle x2 + y2
= 5 at the point (1,-2 ) also touches the circle x2 +
y2 -8x +6y + 20 = 0 and find its point of contact

Monik Modi · Accepted Answer

Dear Student,⇒x2+y2=5⇒at point (1,-2)⇒since this is a negative, let's solve for y, would normally have a ± on it, but only worry about the side:⇒x2+y2=5⇒y2=5-x2⇒y=-5-x2⇒the slope of this semi circle can be found by finding the first derivativ, then solving for the x at the specific points in this case x=1⇒dy/dx=x/5-x2⇒solve for x=1⇒dy/dx=1/5-12⇒dy/dx=1/5-1⇒dy/dx=1/4⇒dy/dx=1/2⇒the slop is 1/2 and we have a point (1,-2) so we can determine it's intercept⇒y=mx+b⇒-2=1/21+b⇒-4/2=1/2+b⇒-5/2=b⇒the tangent line at the specified point is:⇒y=(1/2)x-5/2⇒so now the verify the line is also tangent of⇒x2+y2-8x+6y+20=0⇒first we need to find the point of intersection so solve for the system of equations
 we soulde get a single point of intersection if it's tangent:⇒x2+1/2x-5/22-8x+61/2x-5/2+20=0⇒x2+1/4x2-5/2x+25/4-8x+3x-15+20=0⇒let's just multiply everthing by 4 to get rid of the fractions⇒4x2+x2-10x+25-32x+12x-60+80=0⇒5x2-30x+45=0⇒dividing both sides by 5⇒x2-6x+9=0⇒that is a perfect square trinomial⇒x-32=0⇒x=3⇒there is one x so we can solve y⇒y=(1/2)x-5/2⇒y=(1/2)3-5/2⇒y=3/2-5/2⇒y=-2/2⇒y=-1so there is also one intersection point (-1,3) between the line and the second circle
Regards

Prove that the tangent to the circle x2 + y2 = 5 at the point (1,-2 ) also touches the circle x2 + y2 -8x +6y + 20 = 0 and find its point of contact.

Prove that the tangent to the circle x²+ y² = 5 at the point (1,-2 ) also touches the circle x²+ y²-8x +6y + 20 = 0 and find its point of contact.